Integration Technique Practice: Solving Integrals with Rational Expressions

[jwxie]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I took pictures of my work, please click on the link

First problem

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

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

http://i44.tinypic.com/2s7g7pd.jpg

Compare to the solution, I have an extra 3 and the sqrt(3) at the top...?Thank you

 

[HallsofIvy]

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.