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

Answer

`-(1)/(2)"log"|x+1|+(1)/(4)"log"|x^(2)+1|+(3)/(2)"tan^(-1)x+(x)/(x^(2)+1)+c`

Explanation

`(x^(3)+3x+2)/((x^(2)+1)^(2)(x+1))=(x^(3)+2x+x+2)/((x^(2)+1)^(2)(x+1))`
`=(x(x^(2)+1)+2(x+1))/((x^(2)+1)^(2)(x+1))`
`=(x)/((x^(2)+1)(x+1))+(2)/((x^(2)+1)^(2))`
Again, `(x)/((x^(2)+1)(x+1))=(Ax+B)/((x^(2)+1))+(C)/((x+1))`
`rArr" "x = (Ax + B)(x+1)+C(x^(2)+1)`
On putting x = - 1, we get, `-1 = 2 C rArr C = - 1//2`
ON equation coefficients of `x^(2)`, we get 0 = A + C
`rarr" "A = - C = 1//2`
On putting x = 0, we get 0 = B + C
`rArr" "B = - C = 1//2`
`(x^(3)+3x+2)/((x^(2)+1)^(2)(x+1))=(x+1)/(2(x^(2)+1))-(1)/(2(x+1))+(2)/((x^(2)+1)^(2))`
`therefore I=int(x^(3)+3x+2)/((x^(2)+1)^(2)(x+1))dx`
`=-(1)/(2)int(dx)/(x+1)+(1)/(2)int(x+1)/(x^(2)+1)dx+2int(dx)/((x^(2)+1)^(2))`
`rArr I = - (1)/(2)log|x+1|+(1)/(4)log|x^(2)+1|+(1)/(2)tan^(-1)x+2 I_(1)" "...(i)`
where, `I_(1)=int(dx)/((x^(2)+1)^(2))`
Put `" "x = tan theta`
`rArr" "dx = sec^(2)theta d theta`
`therefore I_(1)=int(sec^(2)theta d theta)/((tan^(2)theta+1)^(2))=intcos^(2)theta d theta =(1)/(2)int(1+cos 2 theta)d theta`
`=(1)/(2)[theta+(1)/(2)sin 2 theta]`
`=(1)/(2)theta + (1)/(2)*(tan theta)/((1 + tan^(2)theta))`
`=(1)/(2)tan^(-1)x+(1)/(2)*(x)/((1+x^(2)))`
From Eq. (i), `I = - (1)/(2)log|x+1|+(1)/(4)log|x^(2)+1|+(3)/(2)tan^(-1)x+(x)/(x^(2)+1)+c`