Evaluate: `int1/((x^2+2x+2)^2)dx`

Explanation

`I=(1)/((x^(2)+2x+2)^(2))dx`
`=(1)/([(x+1)^(2)+1^(2)]^(2))dx`
Put `x+1=tan theta`
or `dx=sec^(2)theta d theta`
` :. I= int(1)/((tan^(2)theta+1)^(2))sec^(2)theta d theta`
`=int cos^(2)theta d theta`
`=(1)/(2)int (1+cos2theta)d theta`
`=(1)/(2)(theta+(sin 2theta)/(2))+C `
`=(1)/(2)(theta +sin theta cos theta)+C`
`=(1)/(2){tan^(-1)(x+1)+(x+1)/(x^(2)+2x+2)}+C`