Difficult Indefinite Integral (substitution problem?)

[Ocasta]

Homework Statement

\int (\frac{x}{\sqrt{x+8}}) dx

The Attempt at a Solution

I got to 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} } = <br /> <br /> <br /> <br /> <br />

 

[rock.freak667]

Try putting u = x+8, that should help you along.

 

[Ray Vickson]

Try putting u = x+8, that should help you along.

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

RGV

 

[Ocasta]

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

RGV

<br /> <br /> u = \sqrt{x+8} \rightarrow du = (1) \frac{1}{2 \sqrt{x+8}} dx<br /> <br />

I don't think that helps...

<br /> <br /> u = x+8 \rightarrow du = 1 dx<br /> <br />

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

 

[gb7nash]

u = x+8 \rightarrow du = 1 dx

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

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

 

[czelaya]

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

 

[dynamicsolo]

<br /> <br /> u = \sqrt{x+8} \rightarrow du = (1) \frac{1}{2 \sqrt{x+8}} dx<br /> <br />

I don't think that helps...

<br /> <br /> u = x+8 \rightarrow du = 1 dx<br /> <br />

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

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.

 

[Ocasta]

I ended up going,

<br /> u = x + 8<br />

<br /> x = u - 8<br />

<br /> \int [u-8]/[sqrt(u)] du<br />

<br /> \int (u-8)u^{-1/2} du<br />

<br /> \int u^{1/2} - 8u^{-1/2} du =<br />

<br /> c + \frac{2}{3} u^{3/2} - 16 u^{1/2}<br />

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

Thanks everybody!

 

[gb7nash]

I ended up going,

u = x + 8
x = 8 - u

Your method is fine, but look at this again.

 

[Ocasta]

Your method is fine, but look at this again.

gb7nash, I went back and fixed that problem.

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