Evaluate: `intx/((x-1)(x^2+4))dx`

Explanation

`(x)/((x-1)(x^(2)+4))dx`
Let `(x)/((x-1)(x^(2)+4))=(A)/(x-1)+(Bx+C)/(x^(2)+4) " " ` (1)
or `x=A(x^(2)+4)+(Bx+C)(x-1) " " `(2)
Putting `x=1` in equation (2), we get `I=5A`.
Putting `x=0` in equation (2), we get `0=4A-C.`
Putting `x= -1` in equation (2), we get `-1=5A+2B-2C`.
Solving these equations, we obtain ` A=(1)/(5), B= -(1)/(5), " and " C=(4)/(5)`.
Substituting the values of A, B, and C in equation (1), we obtain
`(x)/((x-1)(x^(2)+4))=(1)/(5(x-1))+(-(1)/(5)x+(4)/(5))/(x^(2)+4)`
`=(1)/(5(x-1))-(1)/(5)((x-4))/((x^(2)+4))`
or `I=(1)/(5)int(1)/(x-1)dx-(1)/(5)int(x-4)/(x^(2)+4)dx`
`=(1)/(5)int(1)/(x-1)dx-(1)/(10)int(2x)/(x^(2)+4)dx+(4)/(5)int(1)/(x^(2)+4)dx`
`=(1)/(5)log|x-1|-(1)/(10)log(x^(2)+4)+(4)/(5)xx(1)/(2)"tan"^(-1)(x)/(2)+C`
`=(1)/(5)log|x-1|-(1)/(10)log(x^(2)+4)+(2)/(5)"tan"^(-1)((x)/(2))+C`