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#