# Thread: Integration by parts using arctan

1. ## (solved) Integration by parts using arctan

Hi, I'm in need of some help as I can't seem to get this problem right.

Im trying to find the integral of 1/(9x^2+36x+81)dx

So, I started by completing the square and got int 1/[9((x+2)^2+5)] and consequently int 1/(9(x+2)^2+45).

I then turned 9(x+2)^2 into (3(x+2))^2 and factored out the 45 leaving me with (1/45) int 1/[1+(3(x+2))^2 I then used u-sub and made u 3(x+2) and eventually got (1/45)*(1/3) int du/1+u^2.

I know the derivative of arctan is dx/1+x^2 and ended up with

(1/135)arctan(3(x+2)). I know this isn't right as the online program I enter my answer in says its wrong.

If someone could help me out it would be much appreciated!

2. Originally Posted by Phaustus
I then turned 9(x+2)^2 into (3(x+2))^2 and factored out the 45 leaving me with (1/45) int 1/[1+(3(x+2))^2
Here's where you went wrong. It should've been:

$\displaystyle \int \frac{1}{(3(x+2))^2+45}~dx = \int \frac{1}{45((\frac{3x+6}{\sqrt{45}})^2 + 1)}$

3. Ok, yea I knew it was some stupid error like that. I got the correct answer now. Thanks for your help!
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