How do you integrate $(-2x)/((x^2+2x+2)^2)$ ?

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Answer 1 · 2018-04-01T19:55:57

$arctan(x+1)-(x^2-2)/(2x^2+4x+4)+C$

$int (-2x)/(x^2+2x+2)^2*dx$

= $int (-2x)/((x+1)^2+1)^2*dx$

After using $x+1=tanu$ , $x=tanu-1$ and $dx=(secu)^2*du$ transforms, this integral became

$-int 2(tanu-1)*((secu)^2*du)/(secu)^4$

= $-int 2(tanu-1)*(cosu)^2*du$

= $int 2(cosu)^2*du$ - $int 2tanu*(cosu)^2*du$

= $int 2(cosu)^2*du$ - $int 2sinu*cosu*du$

= $int (1+cos2u)*du$ - $int sin2u*du$

= $u+1/2sin2u+1/2cos2u+C$

= $u+1/2*(2tanu)/((tanu)^2+1)+1/2*(1-(tanu)^2)/(1+(tanu)^2)+C$

= $u+tanu/((tanu)^2+1)+1/2*(1-(tanu)^2)/(1+(tanu)^2)+C$

After using $x+1=tanu$ and $u=arctan(x+1)$ inverse transforms, I found

$int (-2x)/(x^2+2x+2)^2*dx$

= $arctan(x+1)+(x+1)/(x^2+2x+2)+1/2*(-x^2-2x)/(x^2+2x+2)+C$

= $arctan(x+1)-(x^2-2)/(2x^2+4x+4)+C$

How do you integrate (-2x)/((x^2+2x+2)^2) (-2x)/((x^2+2x+2)^2) ?