# Completing the square and then using trig substitutio

1. Mar 2, 2009

### KevinL

1. The problem statement, all variables and given/known data
Integral of 1/[(x^2+4x+3)^(3/2)]

3. The attempt at a solution
I tried completing the square and then using trig substitution. So its:

1/[(x+2)^2 -1] let x+2=sec(theta), dx =sec(theta)tan(theta)

After some simplifying, I get it down to integral of csc(theta) which = -ln|csc(theta) + cot(theta)|

My calc teacher has the answers online, and it doesnt have any ln in it at all, so im pretty sure im on the wrong track here. Any suggestions?

2. Mar 2, 2009

### Gib Z

Re: integral

It seems like you completely forgot about the exponent in the denominator.

3. Mar 2, 2009

### swraman

Re: integral

The small mistake you made after completing the square is that you forgit to copy down the ^3/2 in the denominator. It should be:

$$\frac{1}{[(x+2)^{2} -1]^{\frac{3}{2}}}$$.

It becomes much easier when you include it ;). But aside from that mistake you're on teh right general track.

Last edited: Mar 2, 2009