Answer
$-\dfrac{9}{28}$
Work Step by Step
We are given the integral $I=\int_1^2 x(x-2)^{1/3} \ dx$
We will solve the given integral by using u-substitution method.
Let us consider that $u=x-2 \implies dx=du$
So, we can write as: $\int_1^2 x(x-2)^{1/3} \ dx=\int_1^2 (u+2)u^{1/3} \ du\\ = \int_1^2 (u^{4/3}+2u^{1/3}) \ du$
Now, use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Therefore, we have:$ \int_1^2 (u^{4/3}+2u^{1/3}) \ du=[\dfrac{u^{7/3}}{7/3}+\dfrac{u^{4/3}}{4/3}]_1^2+C$
or, $=[\dfrac{3(x-2)^{7/3}}{7}+\dfrac{3(x-2)^{4/3}}{4}]_1^2+C$
or, $=[\dfrac{3(2-2)^{7/3}}{7}+\dfrac{3(2-2)^{4/3}}{4}]-[\dfrac{3(1-2)^{7/3}}{7}+\dfrac{3(1-2)^{4/3}}{4}]$
or, $=-\dfrac{9}{28}$
