Improper Integral Help: Solving \int\frac{1}{\sqrt[3]{x-1}}

[spacetime24]

Homework Statement

Solve the integral $$\int$$$$\frac{1}{\sqrt[3]{x-1}}$$. Upper limit of integration is 1 while lower limit is 0.

Homework Equations

N/A.

The Attempt at a Solution

The only thing that I'm sure about is that the antiderivative of the integral is $$\frac{3}{2}$$(x-1)^(2/3) + C. I know that i need to take the limit of the integral, but I am not sure what the limit should be approaching. 1 Maybe? Since f(x) DNE there. Since I'm stuck on that, I'm kinda stuck on everything else besides the antiderivative.

Any help would be great! Thanks.

 

[Mark44]

This is what you want:
$$\lim_{b \to 1^-} \int_0^b \frac{dx}{\sqrt[3]{x-1}}$$

For a definite integral you don't need the constant of integration.