#Jameel_Academy_Mathematics

1. How To Integral an Equation
Follow Me Step By Step
Can I integral The equation ?
Why I can’t Integral The Equation ?

2. First, you have to memories this formula :
1

3. 2
First, you have to memories this formula :

4. You may know the long division of polynomials .
𝑥2 + 3
𝑥 + 1 𝑥 + 1 𝑥2
+ 3
𝑥
𝑥2
+1𝑥−
−1𝑥
−1
−1𝑥 −1
−
+1
If you find a number
here it means the
long division is
complete .
The final result will be :
𝑥 − 1 +
1
𝑥 + 1
= 𝑥2 + 3
𝑥 + 1
May be the integration of final
result equation is easier than
first equation
..

5. After you memories all integration’s
formula and the previous formula
We will start how to integral the equation

6. Solving
Integration’s
equation
By Basic
Integration’s
Formula
By adding a number or
factor trinomial then
solve by a method or
basic integrationBy method of
integration
1
2
3

7. We will start how to think or know this
question of integral can be solved with
basic or method or by adding a number or
factor trinomials
We Have Some Step Just
Follow It And solve the
equations

8. Steps To Integral The Equation (3-Steps)
Integral the equation with the basic Integration’s FormulaStep One :
All Trigonometric Formula
hyperbolic trigonometric Formula
inverse hyperbolic trigonometric Formula
inverse trigonometric Formula

9. Sometimes the integration question doesn’t show to know this integration
can be solve by basic formulas or not . You need to do some trick to show it
.
Ex :
𝑒 𝑥 𝑑𝑥
(𝑒 𝑥)
1
2 𝑑𝑥
𝑒
𝑥
2 𝑑𝑥
2
1
2
𝑒
𝑥
2 𝑑𝑥
2
1
2
𝑒
𝑥
2 𝑑𝑥
Now we can show or we know this question
is solved by basic formula
2 𝑒
𝑥
2+C
2 𝑒 𝑥 +C

10. If you couldn’t solve the equations
with the basic formulas
Follow The Step Two

11. Integral The Equation With The Method Of IntegrationStep Two :
Integration by Substitution
F(x)
H x × G(x)
=
A
H(x)
+
B
G(x)
𝐹(𝑥)
𝑥2 + 𝑥3 + x
𝑥 ∓ 𝑎 𝑥 ∓ 𝑎
Integration by Trigonometric Substitution
Integration by Parts Integrationby Partial Fractions
For :

12. Again, If you couldn’t solve the
equation with the basic formulas
Follow The Step three

13. Integral The Equation By Adding a number or factor trinomialsStep Three :
Add a number or Use
factor trinomials (Do
Some Trick), as you
like
Use an method of
integration or basic
integration’s formula
then
Ex : 1
𝑒 𝑥 + 1
𝑑𝑥
1 + 𝑒 𝑥
− 𝑒 𝑥
𝑒 𝑥 + 1
𝑑𝑥
1+𝑒 𝑥
𝑒 𝑥+1
dx +
−𝑒 𝑥
𝑒 𝑥 + 1
𝑑𝑥
1
𝑑𝑥 −
𝑒 𝑥
𝑒 𝑥 + 1
𝑑𝑥
f(x)
f’(x)
X-ln|𝑒 𝑥
+1|+c
Note: this Question can
be solved by u-
Substitutionbefore
you solve by this
method

14. Let’s solve an examples
Brief:
 Integration by basic.
 Integration by method.
 Add a number or other trick then Integration by basics or
method

15. e x
x
dx
1
x
e x
dx
2
1
2 x
e x
dx
2 e x
+C
ex−2
x3 cos2 2ex−2 dx
f x
f′ x
1
2 × −2
sec2 2ex−2
2 × −2 x−3 ex−2
dx
f x f′ x
1
2 × −2
× tan 2ex−2
+ c

16. Sinx Cosx dx
1
2
2 𝑆𝑖𝑛𝑥 𝐶𝑜𝑠𝑥 𝑑𝑥
1
2
𝑆𝑖𝑛 2𝑥 𝑑𝑥
-
1
2
cos 2𝑥 + 𝐶
𝟐 𝐒𝐢𝐧𝐱 𝐂𝐨𝐬𝐱 = 𝐒𝐢𝐧(𝟐𝐱)
Use This Formula :
2
9 − 𝑥2
𝑑𝑥
2
1
9(1 −
𝑥2
9
)
𝑑𝑥
2 × 3
9
1
3
1 − (
𝑥
3
)2
𝑑𝑥
2 sin−1
𝑥
3
+ 𝐶

17. sin4 𝑥 𝑑𝑥
(sin2 𝑥)2 𝑑𝑥
(
1
2
− 𝐶𝑜𝑠 2𝑥 )2
𝑑𝑥
1
4
− 𝐶𝑜𝑠 2𝑥 + Cos2( 2𝑥) 𝑑𝑥
1
4
𝑑𝑥 + 𝐶𝑜𝑠 2𝑥 𝑑𝑥 + Cos2( 2𝑥) 𝑑𝑥
Use This Formula :
𝐬𝐢𝐧 𝟐 𝐱 =
𝟏
𝟐
− 𝐂𝐨𝐬 𝟐𝐱
1
4
𝑑𝑥 1
4
𝑥
𝐶𝑜𝑠 2𝑥 𝑑𝑥 1
2
sin(2𝑥)
Cos2( 2𝑥) 𝑑𝑥
Use This Formula :
𝐂𝐨𝐬 𝟐 𝐱 =
𝟏
𝟐
+ 𝐂𝐨𝐬 𝟐𝐱
𝐂𝐨𝐬 𝟐
(𝟐𝐱) =
𝟏
𝟐
+ 𝐂𝐨𝐬 𝟒𝐱
1
2
+ Cos 4x dx
1
2
𝑑𝑥
1
2
𝑥
𝐶𝑜𝑠 4𝑥 𝑑𝑥 1
4
sin(4𝑥)
1
4
𝑥+
1
2
sin 2𝑥 +
1
4
sin 4𝑥 + 𝐶

18. sec2(x)
(1 + tanx)3
dx
sec2
x (1 + tanx)−3
dx
f xf′
x
1
−2
(1 + tan x)−2
+ C
1
−2 (1+tanx)2 +C
ex
e2x + 1
dx
ex
(ex)2 + 1
dx
tan−1
ex
+ C
f x
f′ x

19. 𝑥2
𝑥2 + 1
𝑑𝑥
𝑥2
𝑥2 + 1
𝑑𝑥
𝑥2
𝑥2 + 1
𝑑𝑥
𝑥2 + 1 − 1
𝑥2 + 1
𝑑𝑥
𝑥2 + 1
𝑥2 + 1
𝑑𝑥 +
−1
𝑥2 + 1
𝑑𝑥
𝑑𝑥 −
1
𝑥2 + 1
𝑑𝑥
𝑥 − tan−1
𝑥 + 𝐶
𝑢 = 𝑥2
+ 1
𝑥 = 𝑢 − 1
𝑑𝑥 =
1
2 𝑢 − 1
𝑑𝑢𝑢 − 1
2
𝑢
×
1
2 𝑢 − 1
𝑑𝑢
𝑢 − 1
2𝑢
𝑑𝑢

20. 𝑥 + 1
𝑥 − 1
𝑑𝑥
This Question Can’t be solved by basic
but can be solved by method of
substitute (try it) but for learning we
will solving by same trick .. .
𝑥 + 1
(𝑥 − 1)
×
(𝑥 + 1)
(𝑥 + 1)
𝑑𝑥
𝑥 + 1 2
(𝑥 − 1)(𝑥 + 1)
𝑑𝑥
𝑥
𝑥2 − 1
𝑑𝑥
𝑥 + 1
𝑥2 − 1
𝑑𝑥
+
1
𝑥2 − 1
𝑑𝑥
𝑥(𝑥2−1)
−1
2 𝑑𝑥 +
1
𝑥2 − 1
𝑑𝑥
1
2
2𝑥(𝑥2
−1)
−1
2 +
1
𝑥2 − 1
𝑑𝑥
1
2
(𝑥2−1)
−1
2
1
2
+ cos−1
𝑥 + 𝐶
𝑥2 + 1 + cos−1 𝑥 + 𝐶
(𝒂 𝟐
− 𝒃 𝟐
)= (a + b) (a - b)

21. 1 − 𝑡𝑎𝑛2 𝑥
𝑠𝑒𝑐2 𝑥
𝑑𝑥
This equation solved by us substitute but it will be
difficult , now we will solve by some trick.
−
− 1 − 𝑡𝑎𝑛2 𝑥 + 1 − 1
𝑠𝑒𝑐2 𝑥
𝑑𝑥
−
−1 + 𝑡𝑎𝑛2
𝑥 + 1 − 1
𝑠𝑒𝑐2 𝑥
𝑑𝑥
−
−2 + 𝑡𝑎𝑛2
𝑥 + 1
𝑠𝑒𝑐2 𝑥
𝑑𝑥
−
−2 + 𝑠𝑒𝑐2 𝑥
𝑠𝑒𝑐2 𝑥
𝑑𝑥
-
−2
𝑠𝑒𝑐2 𝑥
𝑑𝑥 +
𝑠𝑒𝑐2 𝑥
𝑠𝑒𝑐2 𝑥
𝑑𝑥
- −2𝑐𝑜𝑠2 𝑥 𝑑𝑥 + 𝑑𝑥
−2 𝑐𝑜𝑠2 𝑥 𝑑𝑥
−2
1
2
+ 𝑐𝑜𝑠2𝑥 𝑑𝑥
1
2
𝑠𝑖𝑛2𝑥
We add a minus and add plus one and minus one to
use this formula :
𝒔𝒆𝒄 𝟐 𝒙 = 𝒕𝒂𝒏 𝟐 𝒙 + 𝟏

22. 𝑥2 + 1
𝑥 + 2 1 − 2𝑥
𝑑𝑥
This Question can’t be solved
with basic formula
Now, We will solve this question
by method (u-substitute)
𝑢 = 1 − 2𝑥
𝑥 =
1
2
−
𝑢2
2
𝑑𝑥 = −𝑢 du
(
1
2
−
𝑢2
2
)2+1
1
2
−
𝑢2
2
+ 2 𝑢
. −𝑢 𝑑𝑢
−
1
4
− 2𝑢2
+
𝑢4
4
5
2 −
𝑢2
2
−
1
4
(1 − 8𝑢2
+ 𝑢4
)
1
2
(5 − 𝑢2)
−
1
2
(1 − 8𝑢2
+ 𝑢4
)
(5 − 𝑢2)
5 − 𝑢2 𝑢4
+ 8𝑢2
− 1
-𝑢2
−5𝑢2 + 𝑢4−
13𝑢2
−13
−65 + 13𝑢2
−
65
−𝑢2 − 13 +
65
5 − 𝑢2
𝑑𝑢
65
5 − 𝑢2
𝑑𝑢
65
1
( 5−𝑢)( 5+𝑢)
𝑑𝑢
𝐴
5−𝑢
+
𝐵
5+𝑢
1 = 𝐴 5 + 𝑢 + 𝐵( 5 − 𝑢)
=
𝑖𝑓 𝑢 = 5 𝐴 =
1
2 5
𝑖𝑓 𝑢 = − 5 𝐵 =
−1
2 5
65[
1
2 5
ln| 5 − 𝑢|-
1
2 5
ln 5 + 𝑢 ]
BA
This Integration Will be
Finish but don’t forget
substitute u = 1 − 2x
After The Trick’s Formula This
Part will solve by partial
fraction
(𝒂 𝟐 − 𝒃 𝟐)= (a + b) (a - b)

23. (𝑥12
+ 4𝑥)10
𝑑𝑥
(𝑥(𝑥11
+ 4))10
𝑑𝑥
𝑥10 (𝑥11 + 4)10 𝑑𝑥
𝑥10 (𝑥11 + 4)10 𝑑𝑥
f x
f′ x
1
11
×
1
11
(𝑥11
+ 4)11
+C
1 + 𝑥
1 + 𝑥2
𝑑𝑥
1
1 + 𝑥2
𝑑𝑥 +
1
2
2𝑥
1 + 𝑥2
𝑑𝑥
1
1 + 𝑥2
𝑑𝑥 +
𝑥
1 + 𝑥2
𝑑𝑥
tan−1 𝑥 +
1
2
𝐼𝑛 1 + 𝑥2 + 𝐶

24. 1 + 𝑥
𝑥2 − 2𝑥
𝑑𝑥
1
2
2 + 2𝑥
𝑥2 − 2𝑥
𝑑𝑥
1
2
2 + 2𝑥 + 4 − 4
𝑥2 − 2𝑥
𝑑𝑥
1
2
2𝑥 − 2
𝑥2 − 2𝑥
𝑑𝑥
1
2
2𝑥 − 2 + 4
𝑥2 − 2𝑥
𝑑𝑥
+
1
2
4
𝑥2 − 2𝑥
𝑑𝑥
Part 1
Part 2
1
2
2𝑥 − 2 (𝑥2 − 2𝑥)
−1
2 𝑑𝑥
Part 1
1
2
(𝑥2
− 2𝑥)
1
2
1
2
Part 2
1
2
4
(𝑥 − 1)2−1
𝑑𝑥
1
2
4
𝑥2 − 2𝑥 + 1 − 1
𝑑𝑥
4
2
cos−1(𝑥 − 1)
1
2
(𝑥2 − 2𝑥)
1
2
1
2
+
4
2
cos−1 𝑥 − 1 + 𝑐

25. END
THE