Observe that [tex](x^2+2x+2)'=2x+2[/tex].
You have that [tex]\int \frac{2x}{x^2+2x+2}\; dx=\int \frac{2x+2-2}{x^2+2x+2}\; dx=\int \frac{2x+2}{x^2+2x+2}\; dx+\int \frac{1}{x^2+2x+2}\; dx=I_1+2I_2[/tex].
To compute [tex]I_1[/tex] you set [tex]y=x^2+2x+2[/tex], so [tex]dy=2x+2\; dx[/tex]. Therefore, [tex]I_1=\int \frac{1}{y}\; dy=\ln |y|=\ln |x^2+2x+2|=\ln (x^2+2x+2)[/tex].
For [tex]I_2[/tex] observe that [tex]x^2+2x+2=(x+1)^2+1[/tex]. Let [tex]y=x+1[/tex]. Then [tex]dy=dx[/tex] and [tex]I_2=\int \frac{1}{y^2+1}=\arctan y=\arctan (x+1)[/tex].
So, [tex]\int \frac{2x}{x^2+2x+2}=\ln (x^2+2x+2)+2\arctan (x+1)+c[/tex].
