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**juicysharpie** $\displaystyle \int_0^2 \ \frac{1}{\sqrt[3]{x-1}}\, dx.$

I know that it is undefined at x=1 because the denominator can't be 0, so I did this:

$\displaystyle \lim_{x \rightarrow 1-} \int_0^c \ \frac{1}{\sqrt[3]{x-1}}\, $ + $\displaystyle \lim_{x \rightarrow 1+} \int_c^2 \ \frac{1}{\sqrt[3]{x-1}}\, $

Then I integrated it...

$\displaystyle \lim_{x \rightarrow 1-} \frac{3}{2}\ (x-1)^\frac{2}{3}$$\displaystyle |_0^C$+$\displaystyle \lim_{x \rightarrow 1+} \frac{3}{2}\ (x-1)^\frac{2}{3}$$\displaystyle |_C^2$

Plugging everything in-

$\displaystyle \lim_{x \rightarrow 1-} \frac{3}{2}\ (c-1)^\frac{2}{3}$ - $\displaystyle \frac{3}{2}\ (0-1)^\frac{2}{3}$ + $\displaystyle \lim_{x \rightarrow 1+} \frac{3}{2}\ (2-1)^\frac{2}{3}$ - $\displaystyle \frac{3}{2}\ (c-1)^\frac{2}{3}$

Then I started to find the limit but somehow I got the wrong answer. I got 1.5-1.5+1.5-1.5 for some reason, because I got that the one sided limits go to 1.5. The answer is supposed to be **3**.

Thank you for your help!!