Answer
$-\dfrac{5}{8} (2-x)^{8/5}+c$
Work Step by Step
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$
where $c$ is a constant of proportionality.
Substitute $k=2-x \implies dx =-dk$
Thus,
$k^{3/5} du=[\dfrac{k^{3/5+1}}{3/5+1}]+c$
or, $=-\dfrac{5}{8} k^{8/5}+c$
plug back in: $k=2-x$, we have
$=-\dfrac{5}{8} (2-x)^{8/5}+c$
