# Homework Help: Difficult Indefinite Integral (substitution problem?)

1. Sep 4, 2011

### Ocasta

1. The problem statement, all variables and given/known data
$\int (\frac{x}{\sqrt{x+8}}) dx$

3. The attempt at a solution
I gotta be honest, I don't even really know where to start with this problem. So bear with me as I take a wild stab in the dark. This section was substitution problems, ...

$\frac{x}{\sqrt{x+8}} = \sqrt{ \frac{x^2}{x+8} } =$

2. Sep 4, 2011

3. Sep 4, 2011

### Ray Vickson

Even better: try putting sqrt(x+8) = u.

RGV

4. Sep 4, 2011

### Ocasta

$u = \sqrt{x+8} \rightarrow du = (1) \frac{1}{2 \sqrt{x+8}} dx$

I don't think that helps...

$u = x+8 \rightarrow du = 1 dx$

I don't think that helps, either. Am I missing something obvious?

5. Sep 4, 2011

### gb7nash

Yes. There's an x on the numerator of the integrand. If u = x + 8, x = ...

6. Sep 4, 2011

### czelaya

Like stated above, u=x+8 makes the integral much simpler to solve.

7. Sep 4, 2011

### dynamicsolo

If you apply Mr. Vickson's method, you need to write $u^{2} = x + 8$, which leads to $2u du = dx$; in the numerator, you then replace the remaining x in the numerator with u2 - 8 . More than one approach works for this integral.

8. Sep 4, 2011

### Ocasta

I ended up going,

$u = x + 8$

$x = u - 8$

$\int [u-8]/[sqrt(u)] du$

$\int (u-8)u^{-1/2} du$

$\int u^{1/2} - 8u^{-1/2} du =$

$c + \frac{2}{3} u^{3/2} - 16 u^{1/2}$

$c + \frac{2}{3} (8+x)^{3/2} - 16 (8+x)^{1/2}$

Thanks everybody!

Last edited: Sep 4, 2011
9. Sep 4, 2011

### gb7nash

Your method is fine, but look at this again.

10. Sep 4, 2011

### Ocasta

gb7nash, I went back and fixed that problem.

Hey guys, thanks to all of you! You've been a great help.