Integral of [ (2x+3) / (x^2 +2x + 5)^2 ] dx
How do I begin? Partial fractions?
Nope, partial fractions won't work here. I'm running on too little sleep to figure out why. But trig. substitution will work!
Note that your integral is: $\displaystyle \displaystyle \int \frac {2x + 3}{[(x + 1)^2 + 4]^2}~dx$
So let $\displaystyle \displaystyle x + 1 = 2 \tan \theta$ and your integral becomes:
$\displaystyle \displaystyle \int \frac {4 \tan \theta + 1}{8 \sec^2 \theta}~d \theta$
which is very doable. The rest is left to you. Don't forget to back-substitute when done