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Integrals of Rational Functions

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In this section we will show how to integrate rational functions, that is, functions of the form P(x) Q(x) where P and Q are polynomials. Integrals of Rational Functions
In this section we will show how to integrate rational functions, that is, functions of the form P(x) Q(x) where P and Q are polynomials. Rational Decomposition Theorem Given reduced P/Q where deg P < deg Q, then P/Q = F1 + F2 + .. + Fn where Fi = orA (ax + b)k Ax + B (ax2 + bx + c)k and that (ax + b)k or (ax2 + bx + c)k are factors of Q(x). Integrals of Rational Functions
Therefore finding the integrals of rational functions is reduced to finding integrals of orA (ax + b)k Ax + B (ax2 + bx + c)k Integrals of Rational Functions
Therefore finding the integrals of rational functions is reduced to finding integrals of orA (ax + b)k Ax + B (ax2 + bx + c)k Integrals of Rational Functions The integrals ∫ are straight forward with the substitution method by setting u = ax + b. dx (ax + b)k
Therefore finding the integrals of rational functions is reduced to finding integrals of orA (ax + b)k Ax + B (ax2 + bx + c)k Integrals of Rational Functions The integrals ∫ are straight forward with the substitution method by setting u = ax + b. dx (ax + b)k To integrate ∫ dx, we need to complete the square of the denominator. Ax + B (ax2 + bx + c)k
Integrals of Rational Functions x x2 + 6x + 10Example A. Find ∫ dx
Complete the square of x2 + 6x + 10 – it’s irreducible. Integrals of Rational Functions x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x ) + 10 Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx = ∫ x (x + 3)2 + 1 dx Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 = ∫ x (x + 3)2 + 1 dx substitution Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution = ∫ u – 3 u2 + 1 du Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution = ∫ u – 3 u2 + 1 du = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution = ∫ u – 3 u2 + 1 du = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Hence Integrals of Rational Functions Set w = u2 + 1 substitution Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
Set w = u2 + 1 = 2u substitution dw du Integrals of Rational Functions = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Set w = u2 + 1
Set w = u2 + 1 = 2u substitution dw du du = dw 2u Integrals of Rational Functions = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Set w = u2 + 1
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u Integrals of Rational Functions = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Set w = u2 + 1
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw Integrals of Rational Functions
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(w) – 3 tan-1(x + 3) + c Integrals of Rational Functions
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions 3x x2 + 7x + 10Example: Find ∫ dx
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions Since x2 + 7x + 10 = (x + 2)(x + 5), we can decompose the rational expression. 3x x2 + 7x + 10Example: Find ∫ dx
Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions 3x x2 + 7x + 10Example: Find ∫ dx Since x2 + 7x + 10 = (x + 2)(x + 5), we can decompose the rational expression. Specifically 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions Hence 3x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x2 + 7x + 10So ∫ dx Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x2 + 7x + 10So ∫ dx = ∫ dx -2 (x + 2) + 5 (x + 5) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x2 + 7x + 10So ∫ dx = -2Ln(lx + 2l) + 5Ln(lx + 5l) + c = ∫ dx -2 (x + 2) + 5 (x + 5) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
Integrals of Rational Functions 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2xExample: Find ∫ dx 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, hence Bx3 = 0, 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, hence Bx3 = 0, so B = 0 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, hence Bx3 = 0, so B = 0 So we've1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 1 (x2 + 1)2xExample: Find ∫ dx
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0.
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this 1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this 1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+ Hence Cx2 + x2 = 0 or C = -1
Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this 1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+ Hence Cx2 + x2 = 0 or C = -1 Put it all together 1 (x2 + 1)2x = -x (x2 + 1) + -x (x2 + 1)2 + 1 x
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx =
Integrals of Rational Functions absolute value arguments for Ln: P - 22,36,61-64
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(lxl)
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(lxl) = - ½ ∫ du u + Ln(lxl)½ ∫ du u2–
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(x) = - ½ ∫ du u + Ln(x)½ ∫ du u2– = - ½ Ln(u) + ½ u-1 + Ln(x) + c
Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(x) = - ½ ∫ du u + Ln(x)½ ∫ du u2– = - ½ Ln(u) + ½ u-1 + Ln(x) + c = - ½ Ln(x2 + 1) + + Ln(x) + c1 2(x2 + 1)

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17 integrals of rational functions x

  • 1. In this section we will show how to integrate rational functions, that is, functions of the form P(x) Q(x) where P and Q are polynomials. Integrals of Rational Functions
  • 2. In this section we will show how to integrate rational functions, that is, functions of the form P(x) Q(x) where P and Q are polynomials. Rational Decomposition Theorem Given reduced P/Q where deg P < deg Q, then P/Q = F1 + F2 + .. + Fn where Fi = orA (ax + b)k Ax + B (ax2 + bx + c)k and that (ax + b)k or (ax2 + bx + c)k are factors of Q(x). Integrals of Rational Functions
  • 3. Therefore finding the integrals of rational functions is reduced to finding integrals of orA (ax + b)k Ax + B (ax2 + bx + c)k Integrals of Rational Functions
  • 4. Therefore finding the integrals of rational functions is reduced to finding integrals of orA (ax + b)k Ax + B (ax2 + bx + c)k Integrals of Rational Functions The integrals ∫ are straight forward with the substitution method by setting u = ax + b. dx (ax + b)k
  • 5. Therefore finding the integrals of rational functions is reduced to finding integrals of orA (ax + b)k Ax + B (ax2 + bx + c)k Integrals of Rational Functions The integrals ∫ are straight forward with the substitution method by setting u = ax + b. dx (ax + b)k To integrate ∫ dx, we need to complete the square of the denominator. Ax + B (ax2 + bx + c)k
  • 6. Integrals of Rational Functions x x2 + 6x + 10Example A. Find ∫ dx
  • 7. Complete the square of x2 + 6x + 10 – it’s irreducible. Integrals of Rational Functions x x2 + 6x + 10Example A. Find ∫ dx
  • 8. x2 + 6x + 10 = (x2 + 6x ) + 10 Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 9. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 10. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 11. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx = ∫ x (x + 3)2 + 1 dx Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 12. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 = ∫ x (x + 3)2 + 1 dx substitution Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 13. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 14. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution = ∫ u – 3 u2 + 1 du Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 15. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution = ∫ u – 3 u2 + 1 du = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Hence Integrals of Rational Functions Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 16. x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9 = (x + 3)2 + 1 x x2 + 6x + 10∫ dx Set u = x + 3 x = u – 3 dx = du= ∫ x (x + 3)2 + 1 dx substitution = ∫ u – 3 u2 + 1 du = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Hence Integrals of Rational Functions Set w = u2 + 1 substitution Complete the square of x2 + 6x + 10 – it’s irreducible. x x2 + 6x + 10Example A. Find ∫ dx
  • 17. Set w = u2 + 1 = 2u substitution dw du Integrals of Rational Functions = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Set w = u2 + 1
  • 18. Set w = u2 + 1 = 2u substitution dw du du = dw 2u Integrals of Rational Functions = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Set w = u2 + 1
  • 19. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u Integrals of Rational Functions = ∫ u u2 + 1 du – 3 ∫ 1 u2 + 1 du Set w = u2 + 1
  • 20. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw Integrals of Rational Functions
  • 21. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(w) – 3 tan-1(x + 3) + c Integrals of Rational Functions
  • 22. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions
  • 23. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions 3x x2 + 7x + 10Example: Find ∫ dx
  • 24. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions Since x2 + 7x + 10 = (x + 2)(x + 5), we can decompose the rational expression. 3x x2 + 7x + 10Example: Find ∫ dx
  • 25. Set w = u2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan-1(u) + cdw = ½ Ln(lwl) – 3 tan-1(x + 3) + c = ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c Integrals of Rational Functions 3x x2 + 7x + 10Example: Find ∫ dx Since x2 + 7x + 10 = (x + 2)(x + 5), we can decompose the rational expression. Specifically 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 26. Integrals of Rational Functions Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 27. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 28. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 29. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 30. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 31. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 32. Integrals of Rational Functions Hence 3x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 33. Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x2 + 7x + 10So ∫ dx Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 34. Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x2 + 7x + 10So ∫ dx = ∫ dx -2 (x + 2) + 5 (x + 5) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 35. Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x2 + 7x + 10So ∫ dx = -2Ln(lx + 2l) + 5Ln(lx + 5l) + c = ∫ dx -2 (x + 2) + 5 (x + 5) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)
  • 36. Integrals of Rational Functions 1 (x2 + 1)2xExample: Find ∫ dx
  • 37. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x 1 (x2 + 1)2xExample: Find ∫ dx
  • 38. Integrals of Rational Functions 1 (x2 + 1)2xExample: Find ∫ dx 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators
  • 39. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 1 (x2 + 1)2xExample: Find ∫ dx
  • 40. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E 1 (x2 + 1)2xExample: Find ∫ dx
  • 41. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 1 (x2 + 1)2xExample: Find ∫ dx
  • 42. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, 1 (x2 + 1)2xExample: Find ∫ dx
  • 43. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, 1 (x2 + 1)2xExample: Find ∫ dx
  • 44. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 1 (x2 + 1)2xExample: Find ∫ dx
  • 45. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, 1 (x2 + 1)2xExample: Find ∫ dx
  • 46. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, hence Bx3 = 0, 1 (x2 + 1)2xExample: Find ∫ dx
  • 47. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, hence Bx3 = 0, so B = 0 1 (x2 + 1)2xExample: Find ∫ dx
  • 48. Integrals of Rational Functions 1 (x2 + 1)2x = Ax + B (x2 + 1) + Cx + D (x2 + 1)2 + E x Clear denominators 1 = (Ax + B)(x2 + 1)x +(Cx + D) x +E(x2 + 1)2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B)(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 There is no x4-term, hence Ax4 + x4 = 0, so A = -1 There is no x3-term, hence Bx3 = 0, so B = 0 So we've1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 1 (x2 + 1)2xExample: Find ∫ dx
  • 49. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2
  • 50. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0.
  • 51. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2
  • 52. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this
  • 53. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this 1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+
  • 54. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this 1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+ Hence Cx2 + x2 = 0 or C = -1
  • 55. Integrals of Rational Functions There is no x-term in 1 = -x(x2 + 1)x +(Cx + D)x +1(x2 + 1)2 Hence Dx = 0, or D = 0. So the expression is 1 = -x2 (x2 + 1)+Cx2 +1(x2 + 1)2 Expand this 1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+ Hence Cx2 + x2 = 0 or C = -1 Put it all together 1 (x2 + 1)2x = -x (x2 + 1) + -x (x2 + 1)2 + 1 x
  • 56. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫
  • 57. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1
  • 58. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x
  • 59. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx =
  • 60. Integrals of Rational Functions absolute value arguments for Ln: P - 22,36,61-64
  • 61. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(lxl)
  • 62. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(lxl) = - ½ ∫ du u + Ln(lxl)½ ∫ du u2–
  • 63. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(x) = - ½ ∫ du u + Ln(x)½ ∫ du u2– = - ½ Ln(u) + ½ u-1 + Ln(x) + c
  • 64. Integrals of Rational Functions 1 (x2 + 1)2xTherefore ∫ dx = ∫ -x dx (x2 + 1) + -x dx (x2 + 1)2 + dx x∫ ∫ substitution set u = x2 + 1 du dx = 2x du 2xdx == ∫ -x u du 2x + ∫ -x u2 du 2x + Ln(x) = - ½ ∫ du u + Ln(x)½ ∫ du u2– = - ½ Ln(u) + ½ u-1 + Ln(x) + c = - ½ Ln(x2 + 1) + + Ln(x) + c1 2(x2 + 1)