Question

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

Answer 1

`=ln((x^2+1)/x)+C`

Explanation:

`(x^2-1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)`
Developing
`(x^2-1)/(x(x^2+1))=(A(x^2+1)+ x(Bx+C))/(x(x^2+1)`
So we can compare the coefficients
`x^2-1=A(x^2+1)+ x(Bx+C)`
For `x^2` , `1=A+B` and `-1=A` and `0=C`
Hence `A=-1`and `B=2`
`int((x^2-1)dx)/(x(x^2+1))=intAdx/x+int((Bx+C)dx)/(x^2+1)`
`=int(-1dx)/x+int((2x)dx)/(x^2+1`
`=-lnx +ln(x^2+1)+C`
`=ln((x^2+1)/x)+C`