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

Question

Aurora Alvarado · Accepted Answer

#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#

Axel Alvarez · Answer

undefined To integrate $ \frac{-2x}{(x^2 + 2x + 2)^2} $, use the method of substitution. Let $ u = x^2 + 2x + 2 $. Then, $ du = (2x + 2) dx $.

Substitute $ u $ and $ du $ into the integral:

$$ \int \frac{-2x}{(x^2 + 2x + 2)^2} dx = \int \frac{-1}{u^2} du $$

This integral can be simplified to:

$$ \int \frac{-1}{u^2} du = -\frac{1}{u} + C $$

Now, revert back to the original variable $ x $ by substituting $ u = x^2 + 2x + 2 $:

$$ = -\frac{1}{x^2 + 2x + 2} + C $$

So, the integral of $ \frac{-2x}{(x^2 + 2x + 2)^2} $ is $ -\frac{1}{x^2 + 2x + 2} + C $.