Evaluate: `int(x^2+1)/((x-1)^2(x+3))dx`

Explanation

`I=int(x^(2)+1)/((x-1)^(2)(x+3))dx`
Let `int(x^(2)+1)/((x-1)^(2)(x+3))=(A)/(x-1)+(B)/((x-1)^(2))+(C)/(x+3) " " ` (1)
or `x^(2)+1=A(x-1)(x+3)+B(x+3)+C(x-1)^(2) " " ` (2)
Putting `x-1=0`, i.e., `x=1` in equation (2), we get
`2=4B " or " B=(1)/(2).` Putting `x+3=0`,
i.e., `x= -3` in equation (2), we get `10=16C " or " C=(5)/(8).`
Equating the coefficients of `x^(2)` on both the sides of the identity of equation (2), we get
`I=A+C " or " A=1-C=1-(5)/(8)=(3)/(8)`
Substituting the values of A, B in equation (1), we get
`(x^(2)+1)/((x-1)^(2)(x+3))=(3)/(8)(1)/(x-1)+(1)/(2)(1)/((x-1)^(2))+(5)/(8)(x+3)`
or `I=(3)/(8)int(1)/(x-1)dx+(1)/(2)int(1)/((x-1)^(2))dx+(5)/(8)int(1)/(x+3)dx`
`=(3)/(8)log|x-1|-(1)/(2(x-1))+(5)/(8)log|x+3|+C`