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

*print*Print*list*Cite

### 1 Answer

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

`int(dx)/((x^2+2x+2)^2)=int1/((x+1)^2+1)^2dx`

Let u=x+1 so du=dx

`=int(du)/((u^2+1)^2)`

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(ds)/(sec^4(s))sec^2(s)`

`=intcos^2(s)ds`

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

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

Let p=2s and dp=2ds

`=s/2+1/2intcos(p)(1/2)2dp`

`=s/2+1/4intcos(p)dp`

`=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:

`=((x^2+2x+2)tan^(-1)(x+1)+x+1)/(2(x+1)^2+2)+C`

`=1/2((x+1)/(x^2+2x+2)+tan^(-1)(x+1))+C`

----------------------------------------------------------------

----------------------------------------------------------------

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

`1/4sin(2tan^(-1)(u))=1/4(2sin(tan^(-1)(u))cos(tan^(-1)(u)))`

`=1/2(u/sqrt(u^2+1))(1/sqrt(u^2+1))`

`=u/(2u^2+u)`

`(tan^(-1)(u))/2+u/(2u^2+u)=((u^2+1)tan^(-1)(u)+u)/(2u^2+u)`