Integral of [ (2x+3) / (x^2 +2x + 5)^2 ] dx

How do I begin? Partial fractions?

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- Oct 8th 2010, 04:45 PMVkLIntegral Help
Integral of [ (2x+3) / (x^2 +2x + 5)^2 ] dx

How do I begin? Partial fractions? - Oct 8th 2010, 05:39 PMJhevon
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