# Integration techinque

1. May 5, 2010

### jwxie

1. The problem statement, all variables and given/known data
(Given integral, and solution from Wolfram Alpha)
integral of 4x/(5+2x+x^2)
solution was -2ln |x^2+2x+5| - 2tan-1 (x+1/2) +c

[PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP20419ac3c2hibb2ic3200002fiag16ag9fa2563?MSPStoreType=image/gif&s=10&w=297&h=40 [Broken]

[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP156619ac3641b9d1cd7700002f4hdf127hhag4ch?MSPStoreType=image/gif&s=63&w=255&h=53 [Broken]
2. Relevant equations

3. The attempt at a solution

First problem
Code (Text):
http://i43.tinypic.com/33mv5l4.jpg
^ minor fix for the picture, instead of -4tan I got -tan because I had 1/4 outside the []
But still, my answer is wrong. The solution showed 2ln *** -2tan ****

Second problem

Code (Text):
http://i39.tinypic.com/34i3995.jpg
I knew I could just first expand the square, and then multiple each term by x. But I want to do it this way, but how the solution was wrong?

Third problem

Code (Text):
http://i44.tinypic.com/2s7g7pd.jpg
Compare to the solution, I have an extra 3 and the sqrt(3) at the top...?

Thank you

Last edited by a moderator: May 4, 2017
2. May 5, 2010

### HallsofIvy

Staff Emeritus
I can't see your work but here's what I would do:
$x^2+ 2x+ 5= x^2+ 2x+ 1+ 4= (x+1)^2+ 4$

If we let u= x+1, then $x^2+ 2x+ 5= u^2+ 4$, x= u- 1, and dx= du.

The integral becomes
$$\int\frac{4x}{x^2+ 2x+5} dx= \int \frac{4(u-1)}{u^2+ 4} du$$
$$= 2\int\frac{2u}{u^2+ 4}- \int \frac{1}{1 +(u/2)^2} du$$

Let v= u^2+ 4 in the first integral and do the second as an arctangent.