Question

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

Answer 1

`-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.

Answer 2

`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 !