# What am I doing wrong in this integration problem?

• January 18th 2009, 09:04 PM
Dana_Scully
What am I doing wrong in this integration problem?
Hi, I've been working on this integration problem, but when I upload it to our homework program the computer keeps saying that it's wrong. It's:

$\int \frac{z^5}{\sqrt[3]{9+z^6}} du$

So what I did was:

$u = 9 + z^6$

$\frac{du}{6} = z^5 dz$

$\frac{1}{6}\int u^{\frac{-1}{3}} du$

$\frac{-1}{18 \sqrt[3]{(9+z^6)^4}}$

Do you think it's a syntax problem, or am I making an obvious mistake?
• January 18th 2009, 09:10 PM
Bilbo Baggins
Quote:

Originally Posted by Dana_Scully
Hi, I've been working on this integration problem, but when I upload it to our homework program the computer keeps saying that it's wrong. It's:

$\int \frac{z^5}{\sqrt[3]{9+z^6}} du$

So what I did was:

$u = 9 + z^6$

$\frac{du}{6} = z^5 dz$

$\frac{1}{6}\int u^{\frac{-1}{3}} du$

$\frac{-1}{18 \sqrt[3]{(9+z^6)^4}}$

Do you think it's a syntax problem, or am I making an obvious mistake?

You made a mistake. Your substitution is good, but you went astray. After your substitution, simply treat the entity as a polynomial. Raise the exponent by one and divide by the new exponent. Then, resubstitute the original expression back in. Lastly, the sure-fire proof is to differentiate your answer and see that you get the original expression.
• January 18th 2009, 09:11 PM
Prove It
Quote:

Originally Posted by Dana_Scully
Hi, I've been working on this integration problem, but when I upload it to our homework program the computer keeps saying that it's wrong. It's:

$\int \frac{z^5}{\sqrt[3]{9+z^6}} du$

So what I did was:

$u = 9 + z^6$

$\frac{du}{6} = z^5 dz$

$\frac{1}{6}\int u^{\frac{-1}{3}} du$

$\frac{-1}{18 \sqrt[3]{(9+z^6)^4}}$

Do you think it's a syntax problem, or am I making an obvious mistake?

I think you'll find that $\int{u^{-\frac{1}{3}}\,du} = \frac{3}{2}u^{\frac{2}{3}}$.
• January 18th 2009, 09:15 PM
Aryth
The mistake is somewhat obvious, we'll start at this part:

$\frac 16 \int u^{-\frac 13} ~du$

Now, the integral of $u^n$ is:

$\frac{u^{n+1}}{n+1}$

So, that gives us:

$\frac 16 \left(\frac{u^{\frac{2}{3}}}{\frac{2}{3}}\right)$

Now we simplify:

$\frac 16 \left(\frac{3u^{\frac{2}{3}}}{2}\right)$

$= \frac{u^{\frac{2}{3}}}{4}$

$\frac{\sqrt[3]{(9 + z^6)^2}}{4}$

And there you go.
• January 18th 2009, 09:22 PM
Dana_Scully
Ah, thank you all so much. There goes me confusing derivation and integration again...not sure what's wrong with my head.