How do you integrate int(2x +3)/(x^4-9x^2)2x+3x49x2 using partial fractions?

1 Answer
May 30, 2018

I=1/18ln|x+3|+1/6ln|x-3|-2/9ln|x|+1/(3x)+cI=118ln|x+3|+16ln|x3|29ln|x|+13x+c

Explanation:

Here,

I=int(2x+3)/(x^4-9x^2)dx=int(2x+3)/(x^2(x^2-9))dxI=2x+3x49x2dx=2x+3x2(x29)dx

I=int(2x+3)/(x^2(x-3)(x+3))dxI=2x+3x2(x3)(x+3)dx

Let,

(2x+3)/(x^2(x-3)(x+3))=A/x+B/x^2+C/(x-3)+D/(x+3)2x+3x2(x3)(x+3)=Ax+Bx2+Cx3+Dx+3

2x+32x+3=Ax(x^2-9)+B(x^2-9)+Cx^2(x+3)+Dx^2(x-3)Ax(x29)+B(x29)+Cx2(x+3)+Dx2(x3)

2x+32x+3=x^3(A+C+D)+x^2(B+3C-3D)+x(-9A)-9Bx3(A+C+D)+x2(B+3C3D)+x(9A)9B

"Comparing co efficients of "Comparing co efficients of x^3,x^2,x andx3,x2,xand "constant term :"

A+C+D=0..........to(1)

B+3C-3D=0.......to(2)

-9A=2=>color(red)(A=-2/9to(3)

-9B=3=>color(red)(B=-1/3to(4)

From (1) and (3)color(white)(..........)From (2) and(4)

C+D=2/9to(5)color(white)(.......)3C-3D=1/3=>C-D=1/9to(6)

Adding (5) and (6) we get

C+C=2/9+1/9=>2C=3/9=>C=3/18=>color(red)(C=1/6

From (5) ,we get

1/6+D=2/9=>D=2/9-1/6=(4-3)/18=>color(red)(D=1/18

So,

I=int[(-2/9)/x+(-1/3)/x^2+(1/6)/(x-3)+(1/18)/(x+3)]dx

=-2/9ln|x|-1/3(x^(-2+1)/(-2+1))+1/6ln|x-3|+1/18ln|x+3| + c

=-2/9ln|x|-1/3(x^(-1)/(-1))+1/6ln|x-3|+1/18ln|x+3|+c

=1/18ln|x+3|+1/6ln|x-3|-2/9ln|x|+1/(3x)+c