Question

How do you integrate `int(x+1)/((x^2-2)(3x-5))` using partial fractions?

Answer 1

`int(x+1)/((x^2-2)(3x-5))=int(-8/7x-11/7)/(x^2-2)+int(24/7)/(3x-5)->-8/7intx/(x^2-2)-11/7int1/(x^2-2)+24/7int1/(3x-5)->-4/7ln(x^2-2)-(11sqrt2)/28 ln((x-sqrt2)/(x+sqrt2))+8/7ln(3x-5)`

Explanation:

To integrate we first have to resolve into partial fraction:
`(x+1)/((x^2-2)(3x-5))=(Ax+B)/((x^2-2))+C/(3x-5)`

`x+1=(Ax+B)(3x-5)+C(x^2-2)`

`x+1=3Ax^2+3Bx-5Ax-5B+Cx^2-2C`

`0=3A+C,3B-5A=1,-5B-2C=1`

`A=8/-7,B=-11/7,C=24/7`

`(x+1)/((x^2-2)(3x-5))=(8/-7x+(-11)/7)/((x^2-2))+(24/7)/(3x-5)`
Then integrate each part.