How do you integrate f(x)=(x^2+x)/((x^2-1)(x+3)(x-9)) using partial fractions?

2 Answers
Jun 21, 2018

-1/16*ln|x+3|+3/32ln|x-9|-1/32ln|x-1|+C

Explanation:

Converting your Integrand into partial fractions we get

f(x)=-1/(12*(x-1))+3/(32(x-9))-1/(16*(x+3))
Integrating this we get the result above.

Aug 11, 2018

int(x^2+x)/((x^2-1)(x+3)(x-9))dx=1/22ln(|x-1|)-5/44ln(|x+3|)+3/44ln(|x-9|)+C, C in RR

Explanation:

I=int(x^2+x)/((x^2-1)(x+3)(x-9))dx

=int(xcancel((x+1)))/(cancel((x+1))(x-1)(x+3)(x-9)dx

Consider that x/((x-1)(x+3)(x-9))=A/(x-1)+B/(x+3)+C/(x-9)

x=A(x+3)(x-9)+B(x-1)(x-9)+C(x-1)(x+3)

x=(A+B+C)x^2+(2C-6A-10B)x-27A+9B-3C

By identification :

[(A+B+C),(-6A-10B+2C),(-27A+9B-3C)]=[(0),(1),(0)]

Solving this system, you will find :

A=1/22, B=-5/44, C=3/44

So:

I=int((1/22)/(x-1)-(5/44)/(x+3)+(3/44)/(x-9))dx

=1/22int1/(x-1)dx-5/44int1/(x+3)dx+3/44int1/(x-9)dx

=1/22ln(|x-1|)-5/44ln(|x+3|)+3/44ln(|x-9|)+C, C in RR

\0/ Here's our answer !