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.