Integral $\int_{0} ^{3} \dfrac {1} {(x-1)^{2/3}}dx$ 
Evaluate the Integral $$\int_{0} ^{3} \dfrac {1} {(x-1)^{2/3}}dx$$

I got the answer to be $3(x-1)^{1/3}+C$ by integral techniques, but my friend says that it's false because the integral is improper. I couldn't understand what he meant so can you explain?

One should be wary that the integrand has a singularity at $x=1$ and thus it is wise to split the integral (call it $I$) between the subdomains $(0,1)$ and $(1,3)$, so we have
$$I = \int_0^1 \frac{1}{(x-1)^{2/3}}dx + \int_1^3 \frac{1}{(x-1)^{2/3}}dx.$$
Since we have just removed a single point, nothing really changes. However, the singularity's nature may make these integrals diverge, so it should be helpful to treat them as improper integrals and follow a limit process with the integration limits. To this end, replace $1$ by $\alpha$ on the left-hand side integral and by $\beta$ on the right-hand side term. Then, take limits as $\alpha\to 1_-$ and $\beta\to 1_ +$.
