$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx$ - correctness of the solution The task is to find the improper integral: $$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx.$$
My solution is (the main part):
$$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx = \int \limits_{-1}^1 \frac{1}{(x-1)^{\frac{2}{3}}} dx + \int \limits_{1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx$$
$$
\int \limits_{-1}^1 \frac{1}{(x-1)^{\frac{2}{3}}} dx = \lim \limits_{c \to 1^-} \int_{-1}^c \frac{1}{(x-1)^{\frac{2}{3}}} dx = 
\lim \limits_{c \to 1^-} \Big{[}  \frac{ (x-1)^{\frac{1}{3}}   }{ \frac{1}{3}  }  \Big{]}_{-1}^c = 3  \lim \limits_{c \to 1^-} \Big{[}   (x-1)^{\frac{1}{3}}    \Big{]}_{-1}^c = 
$$
$$
= 3  \lim \limits_{c \to 1^-} \Big{(} (c-1)^{\frac{1}{3}} - (-2)^{\frac{1}{3}}    \Big{)} = - 3 (-2)^{\frac{1}{3}} \approx 1.11012 - 3.27337 \textrm{ i}
$$
Is a complex number a sensible answer here? Shouldn't it be a real number? I am missing something or forgetting about something?

Note that your answer $$(-2)^{\frac{1}{3}} =-1.259921...$$ is a real number not a complex one. 
In general, one of the cube roots of a real number is always real and in this case we just consider the real answer for the integral.
