1. ## Integral

1. K= $\int \frac {(x + 4)^2}{(x^2 + 8x + 17)^{5/2}}~dx$

I know this integral has something to do with partial fractions. I'm pretty sure you can complete the square of the denominator to make things easier, but im not entirely sure how to proceed with this question.

2. Originally Posted by Belowzero78
1. K= $\int \frac {(x + 4)^2}{(x^2 + 8x + 17)^{5/2}}~dx$

I know this integral has something to do with partial fractions. I'm pretty sure you can complete the square of the denominator to make things easier, but im not entirely sure how to proceed with this question.
Note that you have $\int \frac {(x + 4)^2}{[(x + 4)^2 + 1]^{5/2}}~dx$

Now do a trig substitution: $x + 4 = \tan \theta$

3. Ok, i did the substitution now, im getting tan^2(fata)dfata/sec^3(fata)

4. Originally Posted by Belowzero78
Ok, i did the substitution now, im getting tan^2(fata)dfata/sec^3(fata)
it's "theta"

ok, from $\int \frac {\tan^2 \theta}{\sec^3 \theta}~d \theta$, write everything it terms of sine and cosine. it should be ok from there

5. $

\int {sin^2(\theta)cos(\theta)}d\theta
$

6. Originally Posted by Belowzero78
$

\int {sin^2(\theta)cos(\theta)}d\theta
$
good. now you can finish this off with a substitution. and back-substitute to get the answer in terms of $x$

7. So would i set:
$u=cos \theta$
$du=-sin\theta d\theta$

8. Originally Posted by Belowzero78
So would i set:
$u=cos \theta$
$du=-sin\theta d\theta$
no