$\int\left(x^2+x\right)\left(x^{-8}+2 x^{-9}\right)^{1 / 10} d x$ is equal to

$\frac{5}{11}\left(x^2+2 x\right)^{11 / 10}+C$

Let $I=\int\left(x^2+x\right)\left(x^{-8}+2 x^{-9}\right)^{1 / 10} d x$. Then,

$I=\int(x+1)\left(x^2+2 x\right)^{1 / 10} d x=\frac{1}{2} \int\left(x^2+2 x\right)^{1 / 10} d\left(x^2+2 x\right)$

$\Rightarrow I=\frac{1}{2} \times \frac{\left(x^2+2 x\right)^{11 / 10}}{11 / 10}+C=\frac{5}{11}\left(x^2+2 x\right)^{11 / 10}+C$