Completing the square and then using trig substitutio

[KevinL]

Homework Statement

Integral of 1/[(x^2+4x+3)^(3/2)]

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 doesn't have any ln in it at all, so I am pretty sure I am on the wrong track here. Any suggestions?

 

[Gib Z]

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

 

[swraman]

Homework Statement

Integral of 1/[(x^2+4x+3)^(3/2)]

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)

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.