Integrate the rational functions`x/((x^2+1)(x-1)`

Explanation

` int(x)/((x^(2)+1)(x-1))dx`
`" Let " (x)/((x^(2)+1)(x-1))=(A)/(x-1) +(Bx+C)/(x^(2)+1) .....(1)`
`rArr x=A (x^(2) +1) +(Bx+C)(x-1) .....(2)`
x=1 " then " 1=A(2)+0 rArr A=(1)/(2)`
x=0 " then " 0=A-C rArr c=A=(1)/(2)`
Equating the coefficients of `x^(2)`
`0= A+B rArr B=-A =-(1)/(2)`
`:. int (x)/((x^(2)+1)(x-1))dx `
`=int{((1)/(2))/(x-1)+((-(1)/(2)x+(1)/(2)))/(x^(2)+1)}dx`
`=(1)/(2) int(1)/(x-1)dx +(1)/(2) int ((-x+1))/(x^(2)+1)dx`
`=(1)/(2) int (1)/(x-1)dx -(1)/(2) int (x)/(x^(2)-1) dx+(1)/(2) int (1)/(x^(2)+1)dx`
` =(1)/(2) log |x-1|-(1)/(4)I_(1)+(1)/(2) tan^(-1) x+C_(1)......(3)`
`" where " I_(1)=int(2x)/(x^(2)+1)dx" " underset(rArr 2x dx=dt)(" Let " (x^(2)+1)=t)`
`I_(1)=int(dt)/(t)=log |t|+c_(2)`
`=log |x^(2)+1|+C_(2)`
From equation (3),
`int(x)/((x^(2)+1)(x-1))dx =(1)/(2) log |x-1|`
`-(1)/(4) log |x^(2) +1|+(1)/(2) tan^(-1)x+C`
`(C=C_(1) +C_(2))`