Answer
$$F(x) = \frac{ -5}{4x^8} +C$$
Work Step by Step
Given $$f(x) =\frac{10}{x^9}= 10x^{-9}$$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
F(x) &= \frac{ -10}{8}x^{ -8}+C\\
&= \frac{ -5}{4x^8} +C
\end{align*}
To check
\begin{align*}
F'(x) &= \frac{10}{x^9}\\
&=f(x)
\end{align*}