[tex]\dfrac{x^3+2x^2+5x-2}{(x^2+2x+2)^2}=\dfrac{3x-2}{(x^2+2x+2)^2}+\dfrac x{x^2+2x+2}[/tex]
For the remaining integrals, we can complete the square in the denominator:
[tex]x^2+2x+2=(x+1)^2+1[/tex]
and write each numerator as a polynomial [tex]x+1[/tex]:
[tex]3x-2=3(x+1)-5[/tex]
[tex]x=(x+1)-1[/tex]
then substitute [tex]x+1=\tan y[/tex]. Then [tex]\mathrm dx=\sec^2y\,\mathrm dy[/tex], and we have
[tex]\displaystyle\int\left(\dfrac{3\tan t-5}{(\tan^2t+1)^2}+\dfrac{\tan t-1}{\tan^2t+1}\right)\sec^2t\,\mathrm dt[/tex]
[tex]\displaystyle\int\left(\dfrac{3\tan t-5}{\sec^2t}+\tan t-1\right)\,\mathrm dt[/tex]
[tex]\displaystyle\int(3\tan t\cos^2t-5\cos^2t+\tan t-1)\,\mathrm dt[/tex]
[tex]=-\dfrac32\cos^2t-5\left(\dfrac12t+\dfrac14\sin2t\right)-\ln|\cos t|-t+C[/tex]
[tex]=-\dfrac32\cos^2t-\dfrac54\sin2t-\dfrac72t-\ln|\cos t|+C[/tex]
Now [tex]t=\arctan(x+1)[/tex], so the above simplifies to
[tex]=-\dfrac3{2(x^2+2x+2)}-\dfrac{5(x+1)}{2(x^2+2x+2)}-\dfrac72\arctan(x+1)-\ln\dfrac1{\sqrt{x^2+2x+2}}+C[/tex]
[tex]=-\dfrac{5x+8}{2(x^2+2x+2)}-\dfrac72\arctan(x+1)+\dfrac12\ln(x^2+2x+2)+C[/tex]
