I need help with finding the Integral to this Calc question.

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Find `int(dx)/((x^2+2x+2)^2)` :


Let u=x+1 so du=dx


Let u=tan(s) then `du=sec^2(s)ds`

So `(u^2+1)^2=(1+tan^2(s))^2=sec^4(s)` and `s=tan^(-1)(u)`



`=int (1/2+1/2cos(2s))ds`

`=int1/2 ds+int1/2cos(2s)ds`

Let p=2s and dp=2ds



`=s/2+1/4sin(p)+C`  p=2s:

`=s/2+1/4sin(2s)+C`   `s=tan^(-1)(u)`

`=(u^2tan^(-1)(u)+u+tan^(-1)(u))/(2u^2+2)+C`   ** See below** u=1+x:





** `s/2=(tan^(-1)(u))/2`





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