Hi.

I usually check my work with WolframAlpha and for this problem there seems to be a discrepency...

Here's what I've done:

$\displaystyle \int_1^5\frac{dx}{(x-2)^{1/3}}$

Let $\displaystyle u=x-2\Rightarrow{du}=dx$

This means that

$\displaystyle \int_1^5\frac{dx}{(x-2)^{1/3}}=\int_{x=1}^{x=5}\frac{du}{u^{1/3}}$ .

As for the limits with respect to u, we observe that

$\displaystyle x=1\Rightarrow{u=-1}$ and $\displaystyle x=5\Rightarrow{u=3}$.

Finally, evaluating:

$\displaystyle \int_{-1}^3u^{-1/3}du=\frac{3}{2}u^{2/3}|_{-1}^3=\frac{3}{2}[(3)^{2/3}-(-1)^{2/3}]$

Therefore, $\displaystyle \int_1^5\frac{dx}{(x-2)^{1/3}}=\frac{3}{2}(\sqrt[3]{9}-1)$

But WAlpha gets something

different.