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Evaluate the integral. (Use C for the constant of integration. Enter your answer using function notation - use ln(x) instead of ln x.)

1 Answer

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Answer:


\int\limits {(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} \, dx = (3)/(2)\ln(x^2 + 1) + 4√(3)\tan^(-1)((x)/(\sqrt 3) )+ c

Explanation:

Given


\int\limits {(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} \, dx

Required

Integrate

Using partial fraction, we have:


(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} = (Ax+B)/(x^2 + 1) + (Cx + D)/(x^2 + 3)

Take LCM


(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} = ((Ax+B)(x^2 + 3)+ (Cx + D)(x^2 + 1))/((x^2 + 1)(x^2 + 3))

Cancel out the denominators


3x^3 + 4x^2 + 9x + 4} = (Ax+B)(x^2 + 3)+ (Cx + D)(x^2 + 1)

Open brackets


3x^3 + 4x^2 + 9x + 4} = Ax^3+Bx^2 + 3Ax +3B+ Cx^3 + Dx^2 + Cx + D

Collect like terms


3x^3 + 4x^2 + 9x + 4 = Ax^3+ Cx^3+Bx^2+ Dx^2 + 3Ax+ Cx +3B + D

Compare like terms on opposite sides


Ax^3 + Cx^3 = 3x^3
A + C = 3


Bx^2 + Dx^2 = 4x^2
B + D = 4


3Ax + Cx = 9x
3A + C = 9


3B + D = 4

Subtract
B + D = 4 from
3B + D = 4


3B - B + D - D = 4 - 4


2B + 0 = 0


2B = 0


B = 0


B + D = 4


D =4 - B


D =4 - 0


D =4

Subtract
A + C = 3 from
3A + C = 9


3A - A + C - C = 9 - 3


2A = 6


A = 3


A + C = 3


C = 3 - A


C = 3 - 3


C = 0

So, we have:


A = 3
B = 0
C = 0
D =4


(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} = (Ax+B)/(x^2 + 1) + (Cx + D)/(x^2 + 3)


(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} = (3x+0)/(x^2 + 1) + (0*x + 4)/(x^2 + 3)


(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} = (3x)/(x^2 + 1) + (4)/(x^2 + 3)

The integral becomes:


\int\limits {[(3x)/(x^2 + 1) + (4)/(x^2 + 3)]} \, dx

Split:


\int\limits {(3x)/(x^2 + 1) \, dx + \int\limits {(4)/(x^2 + 3)} \, dx

Split


(3)/(2) \int\limits {(2x)/(x^2 + 1) \, dx + 4\int\limits {(1)/(x^2 + 3)} \ dx

Integrate


(3)/(2)\ln(x^2 + 1) + 4√(3)\tan^(-1)(x)/(\sqrt 3) + c

Hence:


\int\limits {(3x^3 + 4x^2 + 9x + 4)/((x^2 + 1)(x^2 +3))} \, dx = (3)/(2)\ln(x^2 + 1) + 4√(3)\tan^(-1)((x)/(\sqrt 3) )+ c

User Gilliduck
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