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Integrales solucionario
1.
Utilizando el metodo
de sustitucion: 1. β« βπ₯ arctan(βπ₯3) 1 + π₯3 ππ₯ SoluciΓ³n: Sea π’ = ππππ‘ππΰ΅«βπ₯3ΰ΅― β ππ’ = 1 1+π₯3 3 2 βπ₯ ππ₯ β 2 3 ππ’ = 1 1+π₯3 βπ₯ ππ₯ Luego, reemplazando: β« βπ₯ arctanΰ΅«βπ₯3ΰ΅― 1 + π₯3 ππ₯ = 2 3 β« π’ ππ’ = 2 3 π’2 2 + π1 = 1 3 (ππππ‘ππαΰΆ₯π₯3 ) α 2 + πΆ β΄ β« βπ₯ + arctanΰ΅«βπ₯3ΰ΅― 1 + π₯3 ππ₯ = 1 3 (ππππ‘ππ αΰΆ₯π₯3 ) α 2 + πΆ
2.
2. β« ππ₯ β(1 +
π₯2) ππ(π₯ + ΰΆ₯1 + π₯2 ) SoluciΓ³n: Utilizando el mΓ©todo de sustituciΓ³n: Sea π’ = ππΰ΅«π₯ + β1 + π₯2 ΰ΅― β ππ’ = 1 π₯+β1+π₯2 α1 + 2π₯ 2β1+π₯2 α ππ₯ ππ’ = 1 π₯ + β1 + π₯2 (1 + π₯ β1 + π₯2 ) ππ₯ ππ’ = 1 π₯ + β1 + π₯2 α β1 + π₯2 + π₯ β1 + π₯2 α ππ₯ ππ’ = β 1 1 + π₯2 ππ₯ ππ‘ππππ§ππππ πππππππππππ ππ ππππππ ππ ππ πππππππππππ: β« ππ₯ β(1 + π₯2) ππΰ΅«π₯ + β1 + π₯2 ΰ΅― = β« ππ₯ ΰΆ₯(1 + π₯2) β ππΰ΅«π₯ + β1 + π₯2 ΰ΅― Sustituimos en la integral: β« ππ₯ ΰΆ₯(1 + π₯2) β ππΰ΅«π₯ + β1 + π₯2 ΰ΅― = β« ππ’ βπ’ = 2βπ’ + πΆ = 2βππ απ₯ + ΰΆ₯1 + π₯2 α + πΆ β΄ β« ππ₯ ΰΆ₯(1 + π₯2) β ππΰ΅«π₯ + β1 + π₯2 ΰ΅― = 2βππ απ₯ + ΰΆ₯1 + π₯2 α + πΆ
3.
3. β« ππ ππ π₯ πππ 4 π₯
β π‘ππ π₯ πππ 3π₯ ππ₯ SoluciΓ³n: πππππππππ ππ πππ‘πππππ: β« ππ ππ π₯ πππ 4 π₯ β π‘ππ π₯ πππ 3π₯ ππ₯ = β« ππ ππ π₯ πππ 4 π₯ πππ 3π₯ ππ₯ β β« π‘ππ π₯ πππ 3π₯ ππ₯ = β« ππ ππ π₯ πππ π₯ ππ₯ β β« π ππ π₯ πππ 4π₯ ππ₯ ππππ ππ πππππππ πππ‘πππππ βππππππ : π’ = π ππ π₯ β ππ’ = πππ π₯ ππ₯ ππππ ππ π πππ’πππ πππ‘πππππ βππππππ : π§ = πππ π₯ β ππ§ = βπ ππ π₯ ππ₯ β βππ§ = π ππ π₯ ππ₯ ππ’π π‘ππ‘π’ππππ ππ ππ πππ‘πππππ: β« ππ ππ π₯ πππ π₯ ππ₯ β β« π ππ π₯ πππ 4π₯ ππ₯ = β« ππ’ ππ’ + β« ππ§ π§4 = β« ππ’ ππ’ + β« π§β4 ππ§ = ππ’ + π1 β π§β3 3 + π2 = ππ’ β 1 3π§3 + πΆ = ππ ππ π₯ β 1 3πππ 3 π₯ + πΆ β΄ β« ππ ππ π₯ πππ 4 π₯ β π‘ππ π₯ πππ 3π₯ ππ₯ = ππ ππ π₯ β 1 3πππ 3 π₯ + πΆ
4.
πππππππππ ππ πππ
πππ‘πππππππ : β« ππ(3π₯2 + 3)π₯ β 4 π5 ππππ‘ππ π₯ 1 + π₯2 ππ₯ = β« ππ(3π₯2 + 3)π₯ 1 + π₯2 ππ₯ + β« β4 π5 ππππ‘ππ π₯ 1 + π₯2 ππ₯ = β« π₯ ππ 3(π₯2 + 1) 1 + π₯2 ππ₯ β 4 β« π5ππππ‘ππ π₯ 1 + π₯2 ππ₯ ππππ ππ πππππππ πππ‘πππππ βππππππ : π’ = ln 3(π₯2 + 1) β ππ’ = 6π₯ 3(π₯2 + 1) ππ₯ ππ’ = 2π₯ (π₯2 + 1) ππ₯ ππ’ 2 = π₯ (π₯2 + 1) ππ₯ ππππ ππ π πππ’πππ πππ‘πππππ βππππππ : π£ = 5arcta n π₯ β ππ£ = 5 1 + π₯2 ππ₯ ππ£ 5 = 1 1 + π₯2 ππ₯ ππ’π π‘ππ‘π’ππππ ππ ππ πππ‘πππππ: β« ππ(3π₯2 + 3)π₯ β 4 π5 ππππ‘ππ π₯ 1 + π₯2 ππ₯ = β« π’ ππ’ 2 β 4 β« ππ£ ππ£ 5 = 1 2 β« π’ ππ’ β 4 5 β« ππ£ ππ£ = 1 2 π’2 2 + π1 β 4 5 ππ£ + π2 = 1 4 π’2 β 4 5 ππ£ + π3 = 1 4 ln2 3(π₯2 + 1) β 4 5 π5 ππππ‘ππ π₯ + πΆ β΄ β« ππ(3π₯2 + 3)π₯ β 4 π5 ππππ‘ππ π₯ 1 + π₯2 ππ₯ = 1 4 ln2 3(π₯2 + 1) β 4 5 π5 ππππ‘ππ π₯ + πΆ 4. β« ππ(3π₯2 + 3)π₯ β 4 πππππ‘ππ π₯ 1 + π₯2 ππ₯ SoluciΓ³n:
5.
5. β« 2π₯3 + π₯2 2π₯2
β π₯ + 4 ππ₯ ππππ ππ π‘ππππππ πππ‘πππππ βππππππ : π’ = 2π₯2 β π₯ + 4 β ππ’ = (4π₯ β 1)ππ₯ (βββ) β« 3π₯ + 4 2π₯2 β π₯ + 4 ππ₯ = β« 3 απ₯ + 4 3 α 2π₯2 β π₯ + 4 ππ₯ = 3 β« π₯ + 4 3 2π₯2 β π₯ + 4 ππ₯ 3 β« π₯ + 4 3 2π₯2 β π₯ + 4 ππ₯ 4 4 = 3 4 β« 4π₯ + 16 3 2π₯2 β π₯ + 4 ππ₯ = 3 4 β« 4π₯ + 16 3 + 1 β 1 2π₯2 β π₯ + 4 ππ₯ = 3 4 β« 4π₯ β 1 + 19 3 2π₯2 β π₯ + 4 ππ₯ SoluciΓ³n: Simplificaremos mediante divisiΓ³n de polinomios: ( 2π₯3 + π₯2 ) βΆ (2π₯2 β π₯ + 4) = π₯ + 1 β( 2π₯3 β π₯2 + 4π₯) 2π₯2 β 4π₯ β( 2π₯2 β π₯ + 4) β3π₯ β 4 Resto Luego, reemplazando: β« 2π₯3 + π₯2 2π₯2 β π₯ + 4 ππ₯ = β« (π₯ + 1 + β3π₯ β 4 2π₯2 β π₯ + 4 ) ππ₯ πππππππππ ππ πππ‘πππππ: β« (π₯ + 1 + β3π₯ β 4 2π₯2 β π₯ + 4 ) ππ₯ = β« π₯ ππ₯ + β« ππ₯ + β« β(3π₯ + 4) 2π₯2 β π₯ + 4 ππ₯ = π₯2 2 + π₯ + π1 + π2 β β« 3π₯ + 4 2π₯2 β π₯ + 4 ππ₯ (ββ) π΄ππ‘ππ ππ ππππππ§ππ ππ π π’π π‘ππ‘π’ππππ ππππππ§πππ πππππππ π‘πππππππ ππ πππ‘πππππ π΄πππππππππππ πππ 4 ππ πππ‘πππππ π¦ ππ’πππ π π’π ππππππ + 1 β 1 para ππππ πππ’ππ ππ ππ’ = (4π₯ β 1)ππ₯ = 3 β« π₯ + 4 3 2π₯2 β π₯ + 4 ππ₯
6.
πππππππππ ππ πππ‘πππππ: 3 4 β« 4π₯
β 1 + 19 3 2π₯2 β π₯ + 4 ππ₯ = 3 4 β« 4π₯ β 1 2π₯2 β π₯ + 4 ππ₯ + 3 4 β« 19 3 2π₯2 β π₯ + 4 ππ₯ = 3 4 β« 4π₯ β 1 2π₯2 β π₯ + 4 ππ₯ + 3 4 19 3 β« ππ₯ 2π₯2 β π₯ + 4 = 3 4 β« 4π₯ β 1 2π₯2 β π₯ + 4 ππ₯ + 19 4 β« ππ₯ 2π₯2 β π₯ + 4 πΈπ ππ πππππππ πππ‘πππππ π π’π π‘ππ‘π’ππππ ππ π£πππππππ ππ (βββ) π¦ ππππ ππ π πππ’πππ πππ‘πππππ πππππππ ππ’ππππππ ππ πππππππ: = 3 4 β« ππ’ π’ + 19 4 β« ππ₯ 2 απ₯2 β π₯ 2 + 2α = 3 4 ln | π’ | + π3 + 19 8 β« ππ₯ π₯2 β π₯ 2 + 2 = 3 4 ln | 2π₯2 β π₯ + 4 | + π3 + 19 8 β« ππ₯ π₯2 β π₯ 2 + 2 + 1 16 β 1 16 = 3 4 ln | 2π₯2 β π₯ + 4 | + π3 + 19 8 β« ππ₯ απ₯ β 1 4α 2 β 1 16 + 2 = 3 4 ln | 2π₯2 β π₯ + 4 | + π3 + 19 8 β« ππ₯ απ₯ β 1 4 α 2 + 31 16 = 3 4 ln| 2π₯2 β π₯ + 4 | + π3 + 19 8 β« ππ₯ απ₯ β 1 4 α 2 + α β31 4 α 2 = 3 4 ln | 2π₯2 β π₯ + 4 | + π3 + 19 8 1 β31 4 ππππ‘ππ ( π₯ β 1 4 β31 4 ) + π4 ππππ£πππππ π (ββ) π¦ ππππππππ§ππππ ππ πππππ’ππππ πππ‘πππππππππ‘π = π₯2 2 + π₯ + π1 + π2 β ( 3 4 ln| 2π₯2 β π₯ + 4 | + π3 + 19 8 1 β31 4 ππππ‘ππ ( π₯ β 1 4 β31 4 ) + π4) = π₯2 2 + π₯ β 3 4 ln| 2π₯2 β π₯ + 4 | β 19 2β31 ππππ‘ππ ( 4 απ₯ β 1 4 α β31 ) + πΆ β΄ β« 2π₯3 + π₯2 2π₯2 β π₯ + 4 ππ₯ = π₯2 2 + π₯ β 3 4 ln| 2π₯2 β π₯ + 4 | β 19 2β31 ππππ‘ππ ( 4 απ₯ β 1 4 α β31 ) + πΆ
7.
β΄ β« arctan(βπ₯) βπ₯ +
2π₯2 + π₯3 ππ₯ = ππππ‘ππ2 βπ₯ + πΆ 6. β« arctan(βπ₯) βπ₯ + 2π₯2 + π₯3 ππ₯ SoluciΓ³n: Sea: π’ = ππππ‘ππΰ΅«βπ₯ ΰ΅― β ππ’ = 1 1+π₯ 1 2βπ₯ ππ₯ β 2ππ’ = 1 (1 + π₯)βπ₯ ππ₯ Trabajamos el denominador de la integral para conseguir el du: Por el mΓ©todo de sustituciΓ³n: Sustituimos en la integral: β« arctan(βπ₯) βπ₯ + 2π₯2 + π₯3 ππ₯ = β« arctan(βπ₯) ΰΆ₯π₯(1 + 2π₯ + π₯2) ππ₯ = β« arctan(βπ₯) βπ₯ β1 + 2π₯ + π₯2 ππ₯ = β« arctan(βπ₯) βπ₯ ΰΆ₯(1 + π₯)2 ππ₯ = β« arctan(βπ₯) βπ₯ (1 + π₯) ππ₯ = β« arctan(βπ₯) βπ₯ (1 + π₯) ππ₯ = β« π’ β 2ππ’ = 2 π’2 2 + πΆ = π’2 + πΆ = ΰ΅«ππππ‘ππ βπ₯ΰ΅― 2 + πΆ = ππππ‘ππ2 βπ₯ + πΆ
8.
7. β« ππ₯ 1 β
π ππ π₯ + πππ π₯ SoluciΓ³n: Usamos la sustituciΓ³n universal o de Weierstrass: π‘ = π‘ππ( π₯ 2 ) πππ π₯ = 1 β π‘2 1 + π‘2 ; π ππ π₯ = 2π‘ 1 + π‘2 Despejando t para encontrar el ππ‘: ππππ‘ππ (π‘) = π₯ 2 β π₯ = 2ππππ‘ππ (π‘) β ππ₯ = 2 1 + π‘2 ππ‘ Sustituimos en la integral: β« ππ₯ 1 β π ππ π₯ + πππ π₯ = β« 2 1 + π‘2 ππ‘ 1 β 2π‘ 1 + π‘2 + 1 β π‘2 1 + π‘2 = β« 2 1 + π‘2 ππ‘ 1 + π‘2 β 2π‘ + 1 β π‘2 1 + π‘2 = β« 2ππ‘ 2 β 2π‘ = 2 β« ππ‘ β 2(π‘ β 1) = β β« ππ‘ (π‘ β 1) = β ln|π‘ β 1| + πΆ = β ln |π‘ππ α π₯ 2 α β 1| + πΆ Importante: Esta sustituciΓ³n se utiliza generalmente para realizar integrales en la cual veamos funciones trigonomΓ©tricas en el denominador β΄ β« ππ₯ 1 β π ππ π₯ + πππ π₯ = β ln |π‘ππ α π₯ 2 α β 1| + πΆ
9.
β« π2π₯ ππ₯ β1 β ππ₯ 8.
Determine la integral mediante el metodo de integracion por partes: SoluciΓ³n: β« π2π₯ ππ₯ β1 β ππ₯ = β« ππ₯ β ππ₯ β1 β ππ₯ ππ₯ Utilizando el mΓ©todo de integraciΓ³n de partes: π’ = ππ₯ β ππ’ = ππ₯ ππ₯ ππ£ = ππ₯ β1 β ππ₯ ππ₯ β π£ = β2(1 β ππ₯ ) 1 2 = β2β1 β ππ₯ Sustituimos: β« ππ₯ β ππ₯ β1 β ππ₯ ππ₯ = ππ₯ β β2β1 β ππ₯ β β« β2β1 β ππ₯ ππ₯ ππ₯ = β2ππ₯ β1 β ππ₯ + 2 β« ππ₯ β1 β ππ₯ ππ₯ Resolvemos mediante el mΓ©todo de sustituciΓ³n la integral: β«Χ¬β¬ ππ₯ β1 β ππ₯ ππ₯ π§ = 1 β ππ₯ β ππ§ = ππ₯ ππ₯ Sustituyendo, se tiene: β« ππ₯ β1 β ππ₯ ππ₯ = β« βπ§ ππ§ = β« π§ 1 2 ππ§ = π§ 3 2 3 2 + πΆ = π§ 3 2 3 2 + πΆ = 2 3 ΰΆ₯π§3 + πΆ = 2 3 ΰΆ₯(1 β ππ₯)3 + πΆ Luego, β« ππ₯ β ππ₯ β1 β ππ₯ ππ₯ = β2ππ₯ β1 β ππ₯ + 2 β« ππ₯ β1 β ππ₯ ππ₯ = β2ππ₯ β1 β ππ₯ + 2 2 3 ΰΆ₯(1 β ππ₯)3 + πΆ = β2ππ₯ β1 β ππ₯ + 4 3 ΰΆ₯(1 β ππ₯)3 + πΆ β« π π π = π π β β« π π π β΄ β« π2π₯ ππ₯ β1 β ππ₯ = β2ππ₯ β1 β ππ₯ + 4 3 ΰΆ₯(1 β ππ₯)3 + πΆ
10.
9. β« ππ π₯ π₯3(ππ
π₯ β 1)3 ππ₯ SoluciΓ³n: Por propiedad de potencias: β« ππ π₯ π₯3(ππ π₯ β 1)3 ππ₯ = β« ππ π₯ ΰ΅«π₯( ππ π₯ β 1)ΰ΅― 3 ππ₯ Utilizando el mΓ©todo de sustituciΓ³n: π’ = π₯( ππ π₯ β 1) β ππ’ = (π₯ β 1 π₯ + πππ₯ β 1) ππ₯ ππ’ = ππ π₯ ππ₯ Sustituimos en la integral: β« ππ π₯ ΰ΅«π₯( ππ π₯ β 1)ΰ΅― 3 ππ₯ = β« ππ’ π’3 = β« π’β3 ππ’ = π’β2 β2 + πΆ = β 1 2 1 π’2 + πΆ = β 1 2 1 ΰ΅«π₯( ππ π₯ β 1)ΰ΅― 2 + πΆ β΄ β« ππ π₯ π₯3(ππ π₯ β 1)3 ππ₯ = β 1 2 1 ΰ΅«π₯( ππ π₯ β 1)ΰ΅― 2 + πΆ
11.
10. β« π2π₯ ππ₯ β1 β
ππ₯ SoluciΓ³n: Utilizando el mΓ©todo de sustituciΓ³n: Sea: π’ = ππ₯ β ππ’ = ππ₯ ππ₯ Sustituimos en la integral: β« π2π₯ ππ₯ β1 β ππ₯ ππ₯ = β« π’ ππ’ β1 β π’ Utilizando nuevamente el mΓ©todo de sustituciΓ³n: Sea: π§ β = 1 β π’ β π’ = 1 β π§2 ππ§ = β1 2β1 β π’ ππ’ β β2ππ§ = 1 β1 β π’ ππ’ Sustituimos en la integral: β« π’ ππ’ β1 β π’ = β« β2(1 β π§ ) 2 ππ§ = β2 β« ππ§ + 2 β« π§2 = β2π§ + 2 π§3 3 + π1 = β2β1 β π’ + 2 3 ΰ΅«β1 β π’ΰ΅― 3 + π2 = β2β1 β π’ + 2 3 ΰΆ₯(1 β π’)3 + π2 = β2β1 β ππ₯ + 2 3 ΰΆ₯(1 β ππ₯)3 + πΆ β΄ β« π2π₯ ππ₯ β1 β ππ₯ ππ₯ = β2β1 β ππ₯ + 2 3 ΰΆ₯(1 β ππ₯)3 + πΆ
12.
SoluciΓ³n: Utilizando el mΓ©todo
de sustituciΓ³n: Sea: π’ = 2 + tan3 π₯ β ππ’ = 3 tan2 π₯ sec2 π₯ ππ₯ Sustituimos: β« 18 tan2 x sec2 x (2 + tan3 π₯) ππ₯ = β« 18ππ’ π’ = 18 β« ππ’ π’ = 18 ln|π’| + πΆ = 18 ln|2 + tan3 π₯| + πΆ β΄ β« 18 tan2 x sec2 x (2 + tan3 π₯) ππ₯ = 18 ln|2 + tan3 π₯| + πΆ 11. β« 18 tan2 x sec2 x (2 + tan3 π₯) ππ₯
13.
SoluciΓ³n: Utilizando el mΓ©todo
de sustituciΓ³n: Sea: π’ = 1 + π ππ2(π₯ β 1) β ππ’ = 2π ππ(π₯ β 1) cos(π₯ β 1) ππ₯ β ππ’ 2 = π ππ(π₯ β 1) cos(π₯ β 1) ππ₯ Sustituimos: 12. β« ΰΆ₯1 + π ππ2(π₯ β 1) π ππ(π₯ β 1) cos(π₯ β 1) ππ₯ β« ΰΆ₯1 + π ππ2(π₯ β 1) π ππ(π₯ β 1) cos(π₯ β 1) ππ₯ = β« βπ’ ππ’ 2 = 1 2 β« βπ’ ππ’ = 1 2 π’ 3 2 3 2 + πΆ = 3 π’ 3 2 + πΆ = 3 ΰ΅«1 + π ππ2(π₯ β 1)ΰ΅― 3 2 + πΆ β΄ β« ΰΆ₯1 + π ππ2(π₯ β 1) π ππ(π₯ β 1) cos(π₯ β 1) ππ₯ = 3 ΰ΅«1 + π ππ2(π₯ β 1)ΰ΅― 3 2 + πΆ
14.
SoluciΓ³n: Utilizando el mΓ©todo
de sustituciΓ³n: Sea: π’ = 5π₯ + 8 β ππ’ = 5ππ₯ β ππ’ 5 = ππ₯ Sustituimos: β« πx β5π₯ + 8 = β« ππ’ βπ’ = β« π’β 1 2ππ’ = π’β 1 2 +1 β 1 2 + 1 + πΆ = π’ 1 2 1 2 + πΆ = 2 π’ 1 2 + πΆ = 2(5π₯ + 8) 1 2 + πΆ β΄ β« πx β5π₯ + 8 = 2(5π₯ + 8) 1 2 + πΆ SoluciΓ³n: Utilizando el mΓ©todo de sustituciΓ³n: Sea: π’ = π¦3 18 β 1 β ππ’ = 3π¦2 18 ππ¦ β 18ππ’ 3 = π¦2 ππ¦ β 6ππ’ = π¦2 ππ¦ Sustituimos: β« π¦2 α π¦3 18 β 1α 5 ππ¦ = β« π’5 6ππ’ = 6 β« π’5 ππ’ = 6 π’5+1 5 + 1 + πΆ = 6 π’6 6 + πΆ = π’6 + πΆ = α π¦3 18 β 1α 6 + πΆ β΄ β« π¦2 α π¦3 18 β 1α 5 ππ¦ = α π¦3 18 β 1α 6 + πΆ 13. β« πx β5π₯ + 8 14. β« π¦2 α π¦3 18 β 1α 5 ππ¦
15.
SoluciΓ³n: Utilizando el mΓ©todo
de sustituciΓ³n: Sea: π’ = 5π₯ + 8 β ππ’ = 5ππ₯ β ππ’ 5 = ππ₯ Sustituimos: β« πx β5π₯ + 8 = β« ππ’ βπ’ = β« π’β 1 2ππ’ = π’β 1 2 +1 β 1 2 + 1 + πΆ = π’ 1 2 1 2 + πΆ = 2 π’ 1 2 + πΆ = 2(5π₯ + 8) 1 2 + πΆ β΄ β« πx β5π₯ + 8 = 2(5π₯ + 8) 1 2 + πΆ SoluciΓ³n: Utilizando el mΓ©todo de sustituciΓ³n: Sea: π’ = 1 + βπ₯ β ππ’ = 1 2 βπ₯ ππ₯ β 2ππ’ = ππ₯ βπ₯ Sustituimos: β« ΰ΅«1 + βπ₯ΰ΅― 3 πx βπ₯ = β« π’3 ππ’ = π’3+1 3 + 1 + πΆ = π’4 4 + πΆ = ΰ΅«1 + βπ₯ΰ΅― 4 4 + πΆ β΄ β« ΰ΅«1 + βπ₯ΰ΅― 3 πx βπ₯ = ΰ΅«1 + βπ₯ΰ΅― 4 4 + πΆ 15. β« πx π₯β9π₯4 β 4 16. β« ΰ΅«1 + βπ₯ΰ΅― 3 πx βπ₯
16.
SoluciΓ³n: Utilizando el mΓ©todo
de sustituciΓ³n: Sea: π’ = ππ₯ + 1 β ππ’ = ππ₯ ππ₯ Sustituimos: β« ππ₯ πππ ππ (ππ₯ + 1)ππ₯ = β« cosec π’ ππ’ β΄ β« ππ₯ πππ ππ (ππ₯ + 1)ππ₯ = β ln|πππ ππ (ππ₯ + 1) + πππ‘ππ (ππ₯ + 1)| + πΆ SoluciΓ³n: Desarrollando el binomio al cuadrado: β«(sec π‘ + tan π‘)2 ππ‘ = β«(sec2 π‘ + 2 sec π‘ tan π‘ + tan2 π‘)ππ‘ β΄ β«(sec π‘ + tan π‘)2 ππ‘ = 2 tan π‘ + 2sec π‘ β π‘ + πΆ 17. β« ππ₯ πππ ππ (ππ₯ + 1)ππ₯ 18. β«(sec π‘ + tan π‘)2 ππ‘ = β ln|πππ ππ π’ + πππ‘ππ π’| + πΆ = β ln|πππ ππ (ππ₯ + 1) + πππ‘ππ (ππ₯ + 1)| + πΆ = β« sec2 π‘ ππ‘ + β« 2 sec π‘ tan π‘ ππ‘ + β« tan2 π‘ππ‘ = β« sec2 π‘ ππ‘ + 2 β« sec π‘ tan π‘ ππ‘ + β«(π ππ2 π‘ β 1) ππ‘ = β« sec2 π‘ ππ‘ + 2 β« sec π‘ tan π‘ ππ‘ + β« π ππ2 π‘ ππ‘ β β« ππ‘ = 2 β« sec2 π‘ ππ‘ + 2 β« sec π‘ tan π‘ ππ‘ β β« ππ‘ = 2 tan π‘ + 2sec π‘ β π‘ + πΆ
17.
πΌππππ‘ππππ πππ‘πππππππ: π ππ2 π₯ +
cos2 π₯ = 1 SoluciΓ³n: Amplificando por el conjugado β΄ β« ππ₯ 1 + π ππ π₯ = tan π₯ + sec π₯ + πΆ β« ππ₯ 1 + π ππ π₯ β ( 1 β π ππ π₯ 1 β π πππ₯ ) = β« 1 β π ππ π₯ 1 β π ππ2 π₯ ππ₯ 19. β« ππ₯ 1 + π ππ π₯ = β« 1 β π ππ π₯ cos2 π₯ ππ₯ = β« 1 cos2 π₯ ππ₯ β β« π ππ π₯ cos2 π₯ ππ₯ = β« sec2 π₯ ππ₯ β β« 1 cos π₯ π ππ π₯ πππ π₯ ππ₯ = β« sec2 π₯ ππ₯ β β« sec π₯ tan π₯ ππ₯ = tan π₯ + sec π₯ + πΆ
18.
SoluciΓ³n: Separando la integral: β« ππ₯ β
1 ππ₯ + 1 ππ₯ = β« ππ₯ ππ₯ + 1 ππ₯ + β« β1 ππ₯ + 1 ππ₯ = β« ππ₯ ππ₯ + 1 ππ₯ β β« 1 ππ₯ + 1 ππ₯ (ββ) Para la primera integral hacemos la sustituciΓ³n: π’ = ππ₯ + 1 β ππ’ = ππ₯ ππ₯ Sustituimos: β« ππ₯ ππ₯ + 1 ππ₯ = β« ππ’ π’ = ln|π’| + π1 = ln|ππ₯ + 1| + π1 Para le segunda integral sumamos y restando ππ₯ en el numerador: β« 1 ππ₯ + 1 ππ₯ = β« 1 + ππ₯ β ππ₯ ππ₯ + 1 ππ₯ = β« 1 + ππ₯ ππ₯ + 1 ππ₯ + β« βππ₯ ππ₯ + 1 ππ₯ = β« ππ₯ β β« ππ₯ ππ₯ + 1 ππ₯ = π₯ + π2 β β« ππ₯ ππ₯ + 1 ππ₯ Para la integral hacemos: π’ = ππ₯ + 1 β ππ’ = ππ₯ ππ₯ Sustituimos: 20. β« ππ₯ β 1 ππ₯ + 1 ππ₯ π₯ + π2 β β« ππ₯ ππ₯ + 1 ππ₯ = π₯ + π2 β β« ππ’ π’ = π₯ + c2 β ln|π’| + π3 = π₯ + c2 β ln|ππ₯ + 1| + π3
19.
πΌππππ‘ππππππ ππππππππππ : πππ ππ π₯
= 1 π ππ π₯ ; sec π₯ = 1 πππ π₯ πΌππππ‘ππππ ππ’πππππππ‘ππ: tan π₯ = π ππ π₯ cos π₯ Luego, reemplazando en (ββ): β« ππ₯ ππ₯ + 1 ππ₯ β β« 1 ππ₯ + 1 ππ₯ = ln|ππ₯ + 1| + π1 β π₯ + π2 + ln|ππ₯ + 1| + π3 = 2 ln|ππ₯ + 1| + x + C β΄ β« ππ₯ β 1 ππ₯ + 1 ππ₯ = 2 ln|ππ₯ + 1| + x + C 21. β« sec3 π₯ πππ ππ π₯ ππ₯ SoluciΓ³n: Reescribimos la integral: Hacemos: π’ = tan π₯ β ππ’ = sec2 π₯ Sustituimos: β΄ β« sec3 π₯ πππ ππ π₯ ππ₯ = tan2 π₯ 2 + πΆ β« sec3 π₯ πππ ππ π₯ ππ₯ = β« π πππ₯ sec x sec2 π₯ ππ₯ = β« π πππ₯ cos π₯ sec2 π₯ ππ₯ = β« tan π₯ sec2 π₯ ππ₯ β« tan π₯ sec2 π₯ ππ₯ = β« u ππ’ = π’1+1 1 + 1 + πΆ = π’2 2 + πΆ = tan2 π₯ 2 + πΆ
20.
π’ = π ππ(ln
π₯) β ππ’ = cos (ln π₯) 1 π₯ ππ₯ = cos(lnx) π₯ ππ₯ ππ£ = ππ₯ β π£ = π₯ β« π ππ (ln π₯)ππ₯ = π₯ π ππ(ln π₯) β β« π₯ cos(ln π₯) π₯ ππ₯ = π₯ π ππ(ln π₯) β β« cos(ln π₯) ππ₯ π’ = πππ (ππ π₯) β ππ’ = βπ ππ (ππ π₯) 1 π₯ ππ₯ = βπ ππ(πππ₯) π₯ ππ₯ ππ£ = ππ₯ β π£ = π₯ β« π π π = π π β β« π π π SoluciΓ³n: Mediante integraciΓ³n por partes: Sea: Luego, Integrando nuevamente por partes: Sea: Sustituimos: Nos queda la integral inicial por lo tanto reemplazando: β« π ππ(πππ₯) ππ₯ = π₯π ππ(πππ₯) β π₯ πππ (πππ₯) β β« π ππ(πππ₯) ππ₯ Despejando la integral, nos queda: 2 β« π ππ(πππ₯) ππ₯ = π₯π ππ(πππ₯) β π₯ πππ (πππ₯) + π1 β΄ β« π ππ(πππ₯) ππ₯ = 1 2 (π₯π ππ(πππ₯) β π₯ πππ (πππ₯)) + πΆ 22. β« π ππ (ln π₯)ππ₯ π₯ π ππ(ππ π₯) β β« πππ (ππ π₯) ππ₯ = π₯π ππ(πππ₯) β π₯ πππ (πππ₯) + β« π₯ βπ ππ(πππ₯) π₯ ππ₯ = π₯π ππ(πππ₯) β π₯ πππ (πππ₯) β β« π ππ(πππ₯) ππ₯
21.
π’ = arctan
π₯ β ππ’ = 1 1 + π₯2 ππ₯ ππ£ = π₯ππ₯ β π£ = π₯2 2 SoluciΓ³n: Separando las integrales: β« α π₯ + 5 ΰΆ₯(π₯ β 1)5 + π₯ arctan π₯α ππ₯ = β« π₯ + 5 ΰΆ₯(π₯ β 1)5 ππ₯ + β« π₯ arctan π₯ ππ₯ La primera integral realizamos mediante sustituciΓ³n, hacemos: π’ = π₯ β 1 β π₯ = π’ + 1 ππ’ = ππ₯ Luego, Luego, β« π₯ + 5 ΰΆ₯(π₯ β 1)5 ππ₯ = β 2(π₯ β 1)β 1 2 β 4(π₯ β 1)β 3 2 + π1 (1) Para la segunda integral utilizamos el mΓ©todo de integraciΓ³n por partes: Sea: 23. β« α π₯ + 5 ΰΆ₯(π₯ β 1)5 + π₯ arctan π₯α ππ₯ β« π₯ + 5 ΰΆ₯(π₯ β 1)5 ππ₯ = β« π’ + 1 + 5 βπ’5 ππ’ = β« π’ + 6 βπ’5 ππ’ = β« π’ βπ’5 ππ’ + β« 6 βπ’5 ππ’ = β« π’1β 5 2ππ’ + 6 β« π’β 5 2ππ’ = β« π’β 3 2ππ’ + 6 β« π’β 5 2ππ’ = π’β 1 2 β 1 2 + 6 π’β 3 2 β 3 2 + π1 = β2π’β 1 2 β 4π’β 3 2 + π1 = β2(π₯ β 1)β 1 2 β 4(π₯ β 1)β 3 2 + π1
22.
Reemplazamos: Luego, β« π₯ arctan
π₯ ππ₯ = π₯2 2 arctan π₯ β 1 2 π₯ + 1 2 arctan π₯ + π2 (2) AsΓ, de (1) y (2) se tiene: SoluciΓ³n: Por el mΓ©todo de sustituciΓ³n: Sea: π’ = ππππ ππ ΰ΅«βπ₯ ΰ΅― β ππ’ = 1 β1βπ₯ 1 2βπ₯ ππ₯ = 1 2ΰΆ₯(1βπ₯)π₯ ππ₯ = 1 2βπ₯βπ₯2 ππ₯ Sustituimos: β΄ β« ππππ ππ(βπ₯ ) βπ₯ β π₯2 ππ₯ = αππππ ππ ΰ΅«βπ₯ ΰ΅―α 2 + πΆ β« π₯ arctan π₯ ππ₯ = π₯2 2 arctan π₯ β 1 2 β« π₯2 1 + π₯2 ππ₯ = π₯2 2 arctan π₯ β 1 2 β« (1 β 1 π₯2 + 1 ) ππ₯ = π₯2 2 arctan π₯ β 1 2 β« ππ₯ + 1 2 β« 1 π₯2 + 1 ππ₯ β« α π₯ + 5 ΰΆ₯(π₯ β 1)5 + π₯ arctanπ₯αππ₯ = β2(π₯ β 1)β 1 2 β 4(π₯ β 1)β 3 2 + π₯2 2 arctan π₯ β 1 2 π₯ + 1 2 arctan π₯ + πΆ 24. β« ππππ ππ(βπ₯ ) βπ₯ β π₯2 ππ₯ β« ππππ ππ(βπ₯ ) βπ₯ β π₯2 ππ₯ = β« 2 ππππ ππ(βπ₯ ) 2βπ₯ β π₯2 ππ₯ = 2 β« π’ ππ’ = 2 π’2 2 + πΆ =αππππ ππ ΰ΅«βπ₯ ΰ΅―α 2 + πΆ
23.
SoluciΓ³n: Mediante el mΓ©todo
de sustituciΓ³n: πππ: π’ = ln(3π₯) β ππ’ = 1 3π₯ 3ππ₯ = ππ₯ π₯ Sustituimos: β« ππ₯ 2π₯ ln (3π₯) = 1 2 β« ππ’ π’ = 1 2 ln|π’| + πΆ = 1 2 ln|ln(3π₯)| +πΆ 25. β« ππ₯ 2π₯ ln (3π₯) β΄ β« ππ₯ 2π₯ ln (3π₯) = 1 2 ln|ln(3π₯)| +πΆ
24.
cos2 π₯ = 1 +
cos (2π₯) 2 SoluciΓ³n: Podemos escribir: Reemplazando: β« cos2 π₯ ππ₯ = β« ( 1 + πππ (2π₯) 2 ) ππ₯ = β« ( 1 2 + πππ (2π₯) 2 ) ππ₯ = β« 1 2 ππ₯ + β« πππ 2π₯ 2 ππ₯ = 1 2 β« ππ₯ + 1 2 β« cos 2π₯ ππ₯ = 1 2 π₯ + π1 + 1 2 β« cos 2π₯ ππ₯ Para la segunda integral realizamos el mΓ©todo de sustituciΓ³n Sea: π’ = 2π₯ β ππ’ = 2ππ₯ β ππ’ 2 = ππ₯ Sustituimos: 1 2 π₯ + π1 + 1 2 β« cos 2π₯ ππ₯ = 1 2 π₯ + π1 + 1 2 β« cos u ππ’ 2 = 1 2 π₯ + π1 + 1 4 β« cos u ππ’ = 1 2 π₯ + π1 + 1 4 π ππ(π’) + π2 = 1 2 π₯ + 1 4 π ππ(2π₯) + πΆ β΄ β« cos2 π₯ ππ₯ = 1 2 π₯ + 1 2 π ππ(2π₯) + πΆ 26. β« cos2 π₯ ππ₯
25.
SoluciΓ³n: πππ ππ πππ‘πππ
ππ π π’π π‘ππ‘π’ππππ: πππ: π’ = 1 β cos π₯ β ππ’ = π ππ π₯ ππ₯ ππ’π π‘ππ‘π’ππππ : β« π ππ π₯ β1 β cos π₯ ππ₯ = β« βπ’ ππ’ = β« π’ 1 2 ππ’ = π’ 1 2 +1 1 2 + 1 + πΆ = π’ 3 2 3 2 + πΆ = 2 3 π’ 3 2 + πΆ = 2 3 (1 β cos π₯) 3 2 + πΆ β΄ β« π ππ π₯ β1 β cos π₯ ππ₯ = 2 3 (1 β cos π₯) 3 2 + πΆ 27. β«Χ¬β¬ π ππ π₯ β1 β cos π₯ ππ₯
26.
28. β« π₯
cos π₯2 ππ₯ SoluciΓ³n: Por el metodo de sustitucion: πππ: π’ = π₯2 β ππ’ = 2π₯ ππ₯ β du 2 = π₯ ππ₯ ππ’π π‘ππ‘π’ππππ : β« π₯ cos π₯2 ππ₯ = β« cos π’ du 2 = 1 2 β« cos π’ ππ’ = 1 2 sen π’ + πΆ = 1 2 π ππ (π₯2) + πΆ β΄ β« π₯ cos π₯2 ππ₯ = 1 2 π ππ (π₯2) + πΆ
27.
29. β«(tan2 π₯ +
cotan2 π₯ + 4)ππ₯ SoluciΓ³n: Separando las integrales: β«(tan2 π₯ + cotan2 π₯ + 4)ππ₯ = β« tan2 π₯ ππ₯ + β« πππ‘ππ2 π₯ ππ₯ + β« 4ππ₯ = β«(sec2 π₯ β 1 )ππ₯ + β«(πππ ππ2 π₯ β 1) ππ₯ + 4 β« ππ₯ = β« sec2 π₯ ππ₯ β β« ππ₯ + β« πππ ππ2 π₯ ππ₯ β β« ππ₯ + 4 β« ππ₯ = β« sec2 π₯ ππ₯ + β« πππ ππ2 π₯ ππ₯ + 2 β« ππ₯ = tan π₯ β πππ‘ππ π₯ + 2π₯ + πΆ β΄ β«(tan2 π₯ + cotan2 π₯ + 4)ππ₯ = tan π₯ β πππ‘ππ π₯ + 2π₯ + πΆ
28.
30. β« 2πππ‘ππ π₯
β 3π ππ2 π₯ π ππ π₯ ππ₯ SoluciΓ³n: Separando las integrales β« 2πππ‘ππ π₯ β 3π ππ2 π₯ π ππ π₯ ππ₯ = β« 2πππ‘ππ π₯ π ππ π₯ β β« 3π ππ2 π₯ π ππ π₯ ππ₯ = 2 β« πππ ππ π₯ πππ‘ππ π₯ β 3 β« π ππ π₯ ππ₯ Las dos integrales son inmediatas, por lo tanto, integrando: 2 β« πππ ππ π₯ πππ‘ππ π₯ β 3 β« π ππ π₯ ππ₯ = 2(βπππ ππ π₯) β 3(β cos π₯) + πΆ = β2πππ ππ π₯ + 3 cos π₯ + πΆ β΄ β« 2πππ‘ππ π₯ β 3π ππ2 π₯ π ππ π₯ ππ₯ = β2πππ ππ π₯ + 3 cos π₯ + πΆ
29.
31. β« (ππ‘ + 1)3 ππ‘ ππ‘ SoluciΓ³n: Desarrollamos
el binomio al cubo: β« (ππ‘ + 1)3 ππ‘ ππ‘ = β« α π3π‘ + 3π2π‘ + 3ππ‘ + 1 ππ‘ α ππ‘ Separando la integral: β« α π3π‘ + 3π2π‘ + 3ππ‘ + 1 ππ‘ α ππ‘ = β« π3π‘ ππ‘ ππ‘ + β« π2π‘ ππ‘ ππ‘ + β« ππ‘ ππ‘ ππ‘ + β« 1 ππ‘ ππ‘ = β« π3π‘βπ‘ ππ‘ + β« π2π‘βπ‘ ππ‘ + β« ππ‘ + β« πβπ‘ ππ‘ = β« π2π‘ ππ‘ + β« ππ‘ ππ‘ + β« ππ‘ + β« πβπ‘ ππ‘ = π2π‘ + ππ‘ + π‘ β πβπ‘ + πΆ β΄ β« α π3π‘ + 3π2π‘ + 3ππ‘ + 1 ππ‘ α ππ‘ = π2π‘ + ππ‘ + π‘ β πβπ‘ + πΆ
30.
32. β« βπ₯
(π₯ + 1 π₯ ) ππ₯ = SoluciΓ³n: β« βπ₯ (π₯ + 1 π₯ ) ππ₯ = β« βπ₯ (π₯ + π₯β1)ππ₯ = β«ΰ΅«βπ₯ β π₯ + βπ₯ β π₯β1 ΰ΅―ππ₯ = β« π₯ 3 2 ππ₯ + β« π₯β 1 2 ππ₯ = π₯ 3 2 +1 3 2 + 1 + π₯β 1 2 +1 β 1 2 + 1 + πΆ = π₯ 5 2 5 2 + π₯ 1 2 1 2 + πΆ = 2 5 π₯ 5 2 + 2π₯ 1 2 + πΆ β΄ β« βπ₯ (π₯ + 1 π₯ ) ππ₯ = 2 5 π₯ 5 2 + 2π₯ 1 2 + πΆ
31.
33. β«(2π₯ +
1)3 ππ₯ SoluciΓ³n: Desarrollamos el binomio al cubo: β«(2π₯ + 1)3 ππ₯ = β«(8π₯3 + 12π₯2 + 6π₯ + 1) ππ₯ Separando la integral: β«(8π₯3 + 12π₯2 + 6π₯ + 1) ππ₯ = β« 8π₯3 ππ₯ + β« 12π₯2 ππ₯ + β« 6π₯ ππ₯ + β« ππ₯ = 8 β« π₯3 ππ₯ + 12 β« π₯2 ππ₯ + 6 β« π₯ ππ₯ + β« ππ₯ = 8 π₯4 4 + 12 π₯3 3 + 6 π₯2 2 + π₯ + πΆ = 2π₯4 + 4π₯3 + 3π₯2 + π₯ + πΆ β΄ β«(2π₯ + 1)3 ππ₯ = 2π₯4 + 4π₯3 + 3π₯2 + π₯ + πΆ
32.
β« 5π‘2 + 7 π‘ 4 3 dπ‘ =
β« 5π‘2 π‘ 4 3 dπ‘ + β« 7 dπ‘ π‘ 4 3 34. β« 5π‘2 + 7 π‘ 4 3 dπ‘ SoluciΓ³n: = 5 β« π‘2β 4 3 dπ‘ + 7 β« π‘β 4 3ππ‘ = 5 β« π‘ 2 3 dπ‘ + 7 β« π‘β 4 3ππ‘ = 5 π‘ 2 3+1 2 3 + 1 + 7 π‘β 4 3+1 β 4 3 + 1 + πΆ = 5 π‘ 5 3 5 3 + 7 π‘β 1 3 β 1 3 + πΆ = 3 π‘ 5 3 β 21 1 π‘ 1 3 + πΆ β΄ β« 5π‘2 + 7 π‘ 4 3 dπ‘ = 3 π‘ 5 3 β 21 1 π‘ 1 3 + πΆ
33.
35. β« π₯4 πΏπ(3π₯)ππ₯ Solucion: Integrando
por partes, se tiene: π’ = πΏπ(3π₯) β ππ’ = 1 3π₯ 3ππ₯ = 1 π₯ ππ₯ π£ = π₯5 5 β ππ£ = π₯4 β« π₯4 πΏπ(3π₯)ππ₯ = π₯5 5 πΏπ(3π₯) β β« π₯5 5 β 1 π₯ ππ₯ = π₯5 5 πΏπ(3π₯) β 1 5 β« π₯4 ππ₯ = π₯5 5 πΏπ(3π₯) β 1 5 α π₯5 5 α + πΆ = π₯5 5 πΏπ(3π₯) β π₯5 25 + πΆ β« β΄ π₯4 πΏπ(3π₯)ππ₯ = π₯5 5 πΏπ(3π₯) β π₯5 25 + πΆ
34.
β΄ β« π‘ππ6(π₯)ππ₯
= π‘ππ5 (π₯) 5 β π‘ππ3(π₯) 3 + tan(π₯) β π₯ + πΆ 36. β« π‘ππ6(π₯)ππ₯ Solucion: β« π‘ππ6(π₯)ππ₯ = β« π‘ππ2 (π₯) β π‘ππ4(π₯)ππ₯ = β«(π ππ2(π₯) β 1)π‘ππ4 (π₯) ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β« π‘ππ4(π₯) ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β« π‘ππ2(π₯) π‘ππ2(π₯)ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β«(π ππ2(π₯) β 1) π‘ππ2(π₯)ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β« π ππ2(π₯)π‘ππ2(π₯) ππ₯ + β« π‘ππ2(π₯)ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β« π ππ2(π₯)π‘ππ2(π₯) ππ₯ + β« π‘ππ2(π₯)ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β« π ππ2(π₯)π‘ππ2(π₯) ππ₯ + β«(π ππ2(π₯) β 1)ππ₯ = β« π ππ2 (π₯)π‘ππ4 (π₯) ππ₯ β β« π ππ2(π₯)π‘ππ2(π₯) ππ₯ + β« π ππ2(π₯)ππ₯ β β« ππ₯ Para la primera integral hacemos: π’ = tan(π₯) β ππ’ = π ππ2(π₯)ππ₯ Para la segunda integral hacemos: π§ = tan(π₯) β ππ§ = π ππ2(π₯)ππ₯ Sustituimos: = β« π’4 ππ’ β β« π§2 ππ§ + β« π ππ2(π₯)ππ₯ β β« ππ₯ = π’5 5 β π§3 3 + tan(π₯) β π₯ + πΆ = π‘ππ5 (π₯) 5 β π‘ππ3(π₯) 3 + tan(π₯) β π₯ + πΆ
35.
β΄ β« π ππ3(π₯)ππ₯
= β cos(π₯) + cos3 (π₯) 3 + π 37. β« π ππ3(π₯)ππ₯ Solucion: β« π ππ3(π₯)ππ₯ = β« π ππ2 (π₯) β π ππ(π₯) ππ₯ = β« π ππ2 (π₯) β π ππ(π₯) ππ₯ = β«ΰ΅«1 β πππ 2(π₯)ΰ΅― β π ππ(π₯) ππ₯ = β« π ππ(π₯) ππ₯ β β« πππ 2(π₯) β π ππ(π₯)ππ₯ Para la segunda integral hacemos: π’ = cos π₯ β ππ’ = βπ ππ(π₯)ππ₯ = β« π ππ(π₯) ππ₯ + β« πππ 2(π₯) β (βπ ππ(π₯)ππ₯) = β« π ππ(π₯) ππ₯ + β« π’2 ππ’ = β cos(π₯) + π’3 3 + π = β cos(π₯) + cos3 (π₯) 3 + π
36.
β΄ β« ln(cos
π₯) tan π₯ππ₯ = β ln2 (cos π₯) 2 + πΆ β΄ β« ππ₯ cos2 π₯ ΰΆ₯1 + tan π₯ = 2ΰΆ₯1 + tan π₯ + πΆ 38. β« ln(cos π₯) tan π₯ππ₯ Solucion: Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = ln(cos π₯) β ππ’ = βπ ππ π₯ πππ π₯ ππ₯ = β tan π₯ ππ₯ β β ππ’ = tan π₯ ππ₯ Sustituimos: β« ln(cos π₯) tan π₯ ππ₯ = β β« π’ ππ’ = β π’2 2 + πΆ = β ln2 (cos π₯) 2 + πΆ 39. β« ππ₯ cos2 π₯ β1 + tan π₯ Solucion: Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = 1 + tan π₯ β ππ’ = sec2 π₯ ππ₯ β« ππ₯ cos2 π₯ β1 + tan π₯ = β« sec2 π₯ ππ₯ β1 + tan π₯ Sustituimos: β« sec2 π₯ ππ₯ β1 + tan π₯ = β« ππ’ βπ’ = β« π’β 1 2 ππ’ = π’ 1 2 1 2 + πΆ = 2β1 + tan π₯ + πΆ
37.
β΄ β« 1 β
π₯ ln π₯ π₯ ππ₯ ππ₯ = ln π₯ ππ₯ + πΆ β΄ β« π ππ 2π₯ ΰΆ₯1 β cos 2π₯ ππ₯ = 2ΰΆ₯1 β cos 2π₯ + πΆ 40. β« 1 β π₯ ln π₯ π₯ ππ₯ ππ₯ Solucion: Multiplicamos por ππ₯ numerador y denominador β« 1 β π₯ ln π₯ π₯ ππ₯ ππ₯ ππ₯ ππ₯ = β« ππ₯ β ππ₯ π₯ ln π₯ π₯ π2π₯ ππ₯ Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = ln π₯ ππ₯ β ππ’ = ππ₯β 1 π₯ βln π₯ ππ₯ π2π₯ ππ₯ = ππ₯ π₯ β ππ₯ ln π₯ π2π₯ ππ₯ = ππ₯ β ππ₯ ln π₯ π₯ π2π₯ ππ₯ = ππ₯ β ππ₯ ln π₯ π₯ π2π₯ ππ₯ Sustituimos: β« ππ₯ β ππ₯ π₯ ln π₯ π₯ π2π₯ ππ₯ = β« ππ’ = π’ + πΆ = ln π₯ ππ₯ + πΆ 41. β« π ππ 2π₯ β1 β cos 2π₯ ππ₯ Solucion: Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = 1 β cos 2π₯ β ππ’ = π ππ 2π₯ ππ₯ Sustituimos: β« π ππ 2π₯ β1 β cos 2π₯ ππ₯ = β« ππ’ βπ’ = β« π’β 1 2 ππ’ = π’ 1 2 1 2 + πΆ = 2β1 β cos 2π₯ + πΆ
38.
42. β« ππ₯ π ππ2π₯ ΰΆ₯β1
+ πππ‘π π₯ 3 Solucion: Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = πππ‘π π₯ β 1 β ππ’ = βπππ ππ2 π₯ ππ₯ = β 1 π ππ2π₯ ππ₯ β βππ’ = 1 π ππ2π₯ ππ₯ Sustituimos: β« ππ₯ π ππ2π₯ ΰΆ₯β1 + πππ‘π π₯ 3 = β β« ππ’ β π’ 3 = β β« π’β 1 3ππ’ = β π’ 2 3 2 3 + π1 = β 3 2 π’ 2 3 + π1 = β 3 2 (πππ‘π π₯ β 1 ) 2 3 + πΆ β« ππ₯ π ππ2π₯ ΰΆ₯β1 + πππ‘π π₯ 3 = β 3 2 (πππ‘π π₯ β 1 ) 2 3 + πΆ
39.
43. β« π ππ π₯
πtan2 π₯ cos3 π₯ ππ₯ Solucion: Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = πtan2 π₯ β ππ’ = πtan2 π₯ 2 tan π₯ sec2 π₯ ππ₯ β ππ’ 2 = πtan2 π₯ π ππ π₯ cos π₯ 1 cos2 π₯ ππ₯ β ππ’ 2 = πtan2 π₯ π ππ π₯ cos3 π₯ ππ₯ Sustituimos: β« π ππ π₯ πtan2 π₯ ππ₯ cos3 π₯ = 1 2 β« ππ’ = 1 2 π’ + π1 = 1 2 πtan2 π₯ + πΆ β΄ β« π ππ π₯ πtan2 π₯ ππ₯ cos3 π₯ = 1 2 πtan2 π₯ + πΆ
40.
44. β« πarctan π₯ +
π₯ ln(π₯2 + 1) + 1 1 + π₯2 ππ₯ Solucion: Separando la integral: β« πarctan π₯ + ln(π₯2 + 1) + 1 1 + π₯2 ππ₯ = β« πarctan π₯ 1 + π₯2 ππ₯ + β« π₯ ln(π₯2 + 1) 1 + π₯2 ππ₯ + β« 1 1 + π₯2 ππ₯ Para la primera integral hacemos: π’ = πarctan π₯ β ππ’ = 1 1+π₯2 ππ₯ Para la segunda integral hacemos: π§ = ln(π₯2 + 1) β ππ§ = 2π₯ 1+π₯2 ππ₯ β ππ§ 2 = π₯ 1 + π₯2 ππ₯ Sustituimos: β« πarctan π₯ + ln(π₯2 + 1) + 1 1 + π₯2 ππ₯ = β« π’ ππ’ + β« π§ ππ§ 2 + arctan π₯ + π3 = π’2 2 + π1 + 1 2 π§2 2 + π2 + arctan π₯ + π3 = πarctan2 π₯ 2 + 1 4 ln2(π₯2 + 1) + arctan π₯ + πΆ β΄ β« πarctanπ₯ + π₯ ln(π₯2 + 1) + 1 1 + π₯2 ππ₯ = πarctan2 π₯ 2 + 1 4 ln2(π₯2 + 1) + arctan π₯ + πΆ
41.
45. β« π ππ
π₯ ln(1 + π ππ π₯) ππ₯ Solucion: Integrando por partes, se tiene: π’ = ln(1 + π ππ π₯) β ππ’ = cos π₯ 1 + π ππ π₯ ππ₯ π£ = β cos π₯ β ππ£ = π ππ π₯ ππ₯ Luego, β« π ππ π₯ ln(1 + π ππ π₯) ππ₯ = βcos π₯ ln(1 + π ππ π₯) β β« β cos π₯ cos π₯ 1 + π ππ π₯ ππ₯ = βcos π₯ ln(1 + π ππ π₯) + β« cos2 π₯ 1 + π ππ π₯ ππ₯ = βcos π₯ ln(1 + π ππ π₯) + β« 1 β π ππ2 π₯ 1 + π ππ π₯ ππ₯ = βcos π₯ ln(1 + π ππ π₯) + β« (1 + π ππ π₯)(1 β π ππ π₯) 1 + π ππ π₯ ππ₯ = βcos π₯ ln(1 + π ππ π₯) + β«(1 β π ππ π₯) ππ₯ = βcos π₯ ln(1 + π ππ π₯) + β« ππ₯ β β« π ππ π₯ ππ₯ = βcos π₯ ln(1 + π ππ π₯) + π₯ + π1 β (β cos π₯) + π2 = βcos π₯ ln(1 + π ππ π₯) + π₯ + cos π₯ + πΆ β΄ β« π ππ π₯ ln(1 + π ππ π₯) ππ₯ = βcos π₯ ln(1 + π ππ π₯) + π₯ + cos π₯ + πΆ
42.
π§ β2 π₯ ΰΆ₯π₯2 β
2 β2 tan π§ = ΰΆ₯π₯2 β 2 β2 sec π§ = π₯ β ππ₯ = β2 sec π§ tan π§ ππ§ 46. β« ππ₯ (π₯2 β 1)βπ₯2 β 2 Solucion: Mediante sustituciΓ³n trigonomΓ©trica: Sustituimos: β« ππ₯ (π₯2 β 1)βπ₯2 β 2 = β« β2 sec π§ tan π§ ππ§ (2 sec π§ β 1)β2 tan π§ = β« sec π§ ππ§ (2 sec π§ β 1) = β« sec π§ ππ§ 2 (tan2 π§ + 1) β 1 = β« sec π§ ππ§ 2 tan2 π§ + 1 = β« 1 cos π§ ππ§ 2 π ππ2 π§ cos2 π§ + 1 = β« 1 cos π§ ππ§ 2π ππ2 π§ + cos2 π§ cos2 π§ = β« cos π§ ππ§ 2π ππ2 π§ + cos2 π§ = β« cos π§ ππ§ 2π ππ2 π§ + 1 β π ππ2π§ = β« cos π§ ππ§ 1 + π ππ2π§ Sea: π’ = π ππ π§ β ππ’ = cos π§ ππ§
43.
π§ β3 ΰΆ₯(π’ + 2)2
β 3 tan π§ = ΰΆ₯(π’ + 2)2 β 3 β3 β ΰΆ₯(π’ + 2)2 β 3 = β3 tan π§ sec π§ = π’ + 2 β3 β π’ = β3 sec π§ β 2 β ππ’ = β3 sec π§ tan π§ ππ§ Sustituimos β« cos π§ ππ§ 1 + π ππ2π§ = β« ππ’ 1 + π’2 = arctan π’ + πΆ = arctan π ππ π§ + πΆ = arctan α βπ₯2 β 2 π₯ α +πΆ β΄ β« ππ₯ (π₯2 β 1)βπ₯2 β 2 = arctan α βπ₯2 β 2 π₯ α +πΆ 47. β« π ππ π₯ βcos2 π₯ + 4 cos π₯ + 1 ππ₯ Solucion: Mediante el mΓ©todo se sustituciΓ³n: Sea: π’ = cos π₯ β ππ’ = βπ ππ π₯ ππ₯ β βππ’ = π ππ π₯ ππ₯ Sustituimos: β« π ππ π₯ βcos2 π₯ + 4 cos π₯ + 1 ππ₯ = β β« ππ’ βπ’2 + 4π’ + 1 Hacemos completacion de cuadrados: β β« ππ’ βπ’2 + 4π’ + 1 = β β« ππ’ ΰΆ₯(π’ + 2)2 β 3 Utilizamos sustituciΓ³n trigonomΓ©trica: Sustituimos:
44.
β β« ππ’ ΰΆ₯(π’ +
2)2 β 3 = β β« β3 sec π§ tan π§ ππ§ β3 tan π§ = β β« sec π§ ππ§ = β ln | sec π§ + tan π§| + πΆ = β ln | π’ + 2 β3 + ΰΆ₯(π’ + 2)2 β 3 β3 | + πΆ = β ln | cos π₯ + 2 β3 + ΰΆ₯(cos π₯ + 2)2 β 3 β3 | + πΆ β΄ β« π ππ π₯ βcos2 π₯ + 4 cos π₯ + 1 ππ₯ = β ln | cos π₯ + 2 β3 + ΰΆ₯(cos π₯ + 2)2 β 3 β3 | + πΆ 48. β« 3 cos π₯ (π ππ π₯ + 1)(4 β π ππ2π₯) ππ₯ Solucion: Mediante el mΓ©todo se sustituciΓ³n: Sea: π’ = π ππ π₯ β ππ’ = cos π₯ ππ₯ Sustituimos: β« 3 cos π₯ (π ππ π₯ + 1)(4 β π ππ2π₯) ππ₯ = β« 3ππ’ (π’ + 1)(4 β π’2) = β« 3ππ’ (π’ + 1)(2 β π’)(2 + π’) Descomponemos por fracciones parciales: 3 (π’ + 1)(2 β π’)(2 + π’) = π΄ π’ + 1 + π΅ 2 β π’ + πΆ 2 + π’ = π΄(2 β π’)(2 + π’) + π΅(π’ + 1)(2 + π’) + πΆ(π’ + 1)(2 β π’) (π’ + 1)(2 β π’)(2 + π’) Igualando numeradores, se tiene:
45.
Si π’ =
β1 β 3 = π΄ (3)(1) β π΄ = 1 Si π’ = 2 β 3 = π΅(3)(4) β π΅ = 1 4 Si π’ = β2 β 3 = πΆ(β1)(4) β πΆ = β 3 4 Luego, β« 3 cos π₯ (π ππ π₯ + 1)(4 β π ππ2π₯) ππ₯ = β« ( π΄ π’ + 1 + π΅ 2 β π’ + πΆ 2 + π’ ) ππ’ = β« ( 1 π’ + 1 + 1 4 2 β π’ + β 3 4 2 + π’ ) ππ’ = β« ππ’ π’ + 1 + 1 4 β« ππ’ 2 β π’ β 3 4 β« ππ’ 2 + π’ Para la primera integral hacemos: π§ = π’ + 1 β ππ§ = ππ’ Para la segunda integral hacemos: π‘ = 2 β π’ β ππ‘ = βππ’ β βππ‘ = ππ’ Para la tercera integral hacemos: π = 2 + π’ β ππ§ = ππ’ ππ’π π‘ππ‘π’ππππ : = β« ππ’ π§ + 1 4 β« βππ‘ π‘ β 3 4 β« ππ π = ln |π§| + π1 β 1 4 ln |π‘| + π2 β 3 4 ln |π| + π3 = ln |π’ + 1| β 1 4 ln |2 β π’| β 3 4 ln |2 + π’| + πΆ = ln |π ππ π₯ + 1| β 1 4 ln |2 β π ππ π₯| β 3 4 ln |2 + π ππ π₯| + πΆ β΄ β« 3 cos π₯ (π ππ π₯ + 1)(4 β π ππ2π₯) ππ₯ = ln |π ππ π₯ + 1| β 1 4 ln |2 β π ππ π₯| β 3 4 ln |2 + π ππ π₯| + πΆ
46.
π§ 1 tan π§ =
π₯ β 1 β π₯ = tan π§ + 1 β ππ₯ = sec2 π§ ππ§ sec π§ = ΰΆ₯(π₯ β 1)2 + 1 π₯ β 1 49. β« 4π₯ + 5 (π₯2 β 2π₯ + 2) 3 2 ππ₯ Solucion: Hacemos completacion de cuadrados: β« 4π₯ + 5 (π₯2 β 2π₯ + 2) 3 2 ππ₯ = β« 4π₯ + 5 [ΰΆ₯(π₯ β 1)2 + 1] 3 Sustituimos: β« 4π₯ + 5 [ΰΆ₯(π₯ β 1)2 + 1] 3 ππ₯ = β« 4(tan π§ + 1) + 5 [sec π§]3 sec2 π§ ππ§ = β« 4 tan π§ + 9 π ππ π§ ππ§ = 4 β« tan π§ π ππ π§ ππ§ + β« 9 π ππ π§ ππ§ = 4 β« π ππ π§ cos π§ 1 cos π§ ππ§ + 9 β« 1 1 cos π§ ππ§ = 4 β« π ππ π§ ππ§ + 9 β« cos π§ ππ§ = β4 cos π§ + 9π ππ π§ + πΆ = β4 cos 1 ΰΆ₯(π₯ β 1)2 + 1 + 9π ππ π₯ β 1 ΰΆ₯(π₯ β 1)2 + 1 + πΆ β΄ β« 4π₯ + 5 (π₯2 β 2π₯ + 2) 3 2 ππ₯ = β4 cos 1 ΰΆ₯(π₯ β 1)2 + 1 + 9π ππ π₯ β 1 ΰΆ₯(π₯ β 1)2 + 1 + πΆ
47.
50. β« ππ₯ π₯ΰ΅«β1 +
π₯ β 1ΰ΅― Solucion: Mediante el mΓ©todo de sustituciΓ³n: Sea: π’ = β1 + π₯ β π₯ = π’2 β 1 ππ’ = ππ₯ 2β1 + π₯ β 2β1 + π₯ ππ’ = ππ₯ Sustituimos: β« ππ₯ π₯ΰ΅«β1 + π₯ β 1ΰ΅― = β« 2β1 + π₯ ππ’ (π’2 β 1)(π’ β 1) = 2 β« π’ ππ’ (π’2 β 1)(π’ β 1) = 2 β« π’ ππ’ (π’ + 1)(π’ β 1)2 Descomponemos por fracciones parciales: π’ (π’ + 1)(π’ β 1)2 = π΄ π’ + 1 + π΅ π’ β 1 + πΆ (π’ β 1)2 = π΄(π’ β 1)2 + π΅(π’ + 1)(π’ β 1) + πΆ(π’ + 1) (π’ + 1)(π’ β 1)2 Igualando los numeradores, se tiene: Si π’ = β1 β β1 = π΄ (4) β π΄ = β 1 4 Si π’ = 1 β 1 = πΆ (2) β πΆ = 1 2 Si π’ = 0 β 0 = π΄(β1)2 + π΅(1)(β1) + πΆ β π΅ = π΄ + πΆ = β 1 4 + 1 2 β π΅ = 1 4 Luego, π’ (π’ + 1)(π’ β 1)2 = β 1 4 π’ + 1 + 1 4 π’ β 1 + 1 2 (π’ β 1)2 β« ππ₯ π₯ΰ΅«β1 + π₯ β 1ΰ΅― = 2 β« ( β 1 4 π’ + 1 + 1 4 π’ β 1 + 1 2 (π’ β 1)2 ) ππ’
48.
= β 1 2 β« ππ’ π’ +
1 + 1 2 β« ππ’ π’ β 1 + β« ππ’ (π’ β 1)2 Para la primera integral hacemos: π‘ = π’ + 1 β ππ‘ = ππ’ Para la segunda integral hacemos: π = π’ β 1 β ππ = ππ’ Para la tercera integral hacemos: π£ = π’ β 1 β ππ£ = ππ’ = β 1 2 β« ππ‘ π‘ + 1 2 β« ππ π + β« ππ£ π£2 = β 1 2 β« ππ‘ π‘ + 1 2 β« ππ π + β« π£β2 ππ£ = β 1 2 ln |π‘| + π1 + 1 2 ln |π| + π2 + 1 2 π£β1 β1 + π3 = β 1 2 ln |π’ + 1 | + 1 2 ln |π’ β 1 | β 1 (π’ β 1) + π4 = β 1 2 ln |β1 + π₯ + 1 | + 1 2 ln |β1 + π₯ β 1 | β 1 (β1 + π₯ β 1) + πΆ β΄ β« ππ₯ π₯ΰ΅«β1 + π₯ β 1ΰ΅― = β 1 2 ln |β1 + π₯ + 1 | + 1 2 ln |β1 + π₯ β 1 | β 1 (β1 + π₯ β 1) + πΆ 51. β« π₯ ln π₯ ππ₯ β1 β π₯2 Solucion: Integrando por partes, se tiene: π’ = ππ π₯ β ππ’ = 1 π₯ ππ₯ ππ£ = π₯ (1 β x2) 1 2 ππ₯ β π£ = β« π₯ (1 β x2) 1 2 ππ₯ Para hallar v, Hacemos π§ = 1 β π₯2 β ππ§ = β2π₯ππ₯ β β2ππ§ = π₯ππ₯ ππ£ = π₯ (1 β x2) 1 2 ππ₯ β π£ = β« β2ππ§ π§ 1 2 = β2 β« π§β 1 2 ππ’ = β2 π§ 1 2 1 2 = β(1 β x2) 1 2 Luego,
49.
β« π₯ ln π₯
ππ₯ β1 β π₯2 = β(1 β x2) 1 2 ln π₯ β β« β(1 β x2) 1 2 1 π₯ ππ₯ = β(1 β x2) 1 2 ln π₯ + β« β1 β x2 π₯ ππ₯ Sea: π’2 = 1 β π₯2 β π₯2 = 1 β π’2 β π’ = β1 β π₯2 β 2π’ ππ’ = β2π₯ππ₯ β π’ ππ’ = βπ₯ ππ₯ β π’ ππ’ βπ₯ = ππ₯ = β(1 β x2) 1 2 ln π₯ + β« π’ π₯ π’ ππ’ βπ₯ = β(1 β x2) 1 2 ln π₯ β β« π’2 π₯2 ππ’ = β(1 β x2) 1 2 ln π₯ + β« βπ’2 + 1 β 1 1 β π’2 ππ’ = β(1 β x2) 1 2 ln π₯ + β« 1 β π’2 1 β π’2 ππ’ β β« 1 1 β π’2 ππ’ = β(1 β x2) 1 2 ln π₯ + β« ππ’ β β« 1 1 β π’2 ππ’ = β(1 β x2) 1 2 ln π₯ + π’ + π1 β 1 2 ln | π’ + 1 π’ β 1 | + π2 = β(1 β x2) 1 2 ln π₯ + π’ + π1 β 1 2 ln | π’ + 1 π’ β 1 | + π2 = β(1 β x2) 1 2 ln π₯ + ΰΆ₯1 β x2 β 1 2 ln | β1 β x2 + 1 β1 β x2 β 1 | + πΆ β΄ β« π₯ ln π₯ ππ₯ β1 β π₯2 = β(1 β x2) 1 2 ln π₯ + ΰΆ₯1 β x2 β 1 2 ln | β1 β x2 + 1 β1 β x2 β 1 | + πΆ
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