Find : `intx/((x^2+1)(x-1))dx`

Explanation

`(x)/((x-1)(x^(2)+1))=(A)/(x-1)+(Bx+C)/(x^(2)+1)`
`impliesx=A(x^(2)+1)+(Bx+C)(x-1)`
Comparing coefficient of `x^(2)`, `x`, constant term, we get
`{:(A+B=0),(-B+C=1),(A-C=0):}}{:(A=(1)/(2)),(impliesB=(-1)/(2)),(C=(1)/(2)):}`
`:.int(x)/((x-1)(x^(2)+1))dx=int(1)/(2(x-1))dx+int((-1)/(2)x+(1)/(2))/(x^(2)+1)dx`
`=(1)/(2)int(1)/((x-1))dx+(1)/(2)[int(-x)/(x^(2)+1)dx+int(1)/(x^(2)+1)dx]`
`=(1)/(2)int(dx)/(x-1)-(1)/(4)int(2x)/(x^(2)+1)dx+(1)/(2)int(dx)/(x^(2)+1)` put `x^(2)+1=t`
`2xdx=dt`
`=(1)/(2)log|x-1|-(1)/(4)log|x^(2)+1|+(1)/(2)tan^(-1)(x)+C`