How do you integrate (5x)/(2x^2+11x+12)5x2x2+11x+12 using partial fractions?

1 Answer
Nov 24, 2016

int((5x)/(2x^2 + 11x + 12))dx = 4ln|x + 4| - 3/2ln|2x + 3| + C(5x2x2+11x+12)dx=4ln|x+4|32ln|2x+3|+C

Explanation:

We start by factoring the denominator.

2x^2 + 11x + 12 = 2x^2 + 8x + 3x + 12 = 2x(x + 4) + 3(x +4) = (2x + 3)(x + 4)2x2+11x+12=2x2+8x+3x+12=2x(x+4)+3(x+4)=(2x+3)(x+4).

The partial fraction decomposition will therefore be of the form:

A/(2x + 3) + B/(x + 4) = (5x)/((2x + 3)(x + 4)A2x+3+Bx+4=5x(2x+3)(x+4)

A(x + 4) + B(2x + 3) = 5xA(x+4)+B(2x+3)=5x

Ax + 4A + 2Bx+ 3B = 5xAx+4A+2Bx+3B=5x

(A + 2B)x + (4A + 3B) = 5x(A+2B)x+(4A+3B)=5x

We now write a systems of equations.

{(A + 2B= 5), (4A + 3B = 0):}

Solve:

A = 5 - 2B

4(5 - 2B) + 3B = 0

20 - 8B + 3B = 0

-5B = -20

B = 4

A + 2(4) = 5

A = -3

Hence, the partial fraction decomposition is 4/(x + 4) - 3/(2x + 3). The integral becomes

int(4/(x + 4) - 3/(2x + 3))dx

We know that int(1/u)du = ln|u| + C. Therefore:

int(4/(x + 4) - 3/(2x + 3))dx = 4ln|x + 4| - 3/2ln|2x + 3| + C

Hopefully this helps!