Indefinite integral question. Evaluate the following indefinite integral.
$$ \int { \frac { 8 }{ 81+{ x }^{ 2 } } } dx $$
The answer is 
$$ \frac { 8 }{ 9 } \arctan \left(\frac { x }{ 9 } \right)$$
I know that it has something to do with this integral 
$$ \int { \frac { 1 }{ { x }^{ 2 }+1 }  } dx =\arctan x  +C $$
but i can't get it.
I think i can't make the algebra to find the antiderivative.

For
$$\int\frac{8}{81+x^2} \, dx,$$
If we let
$$
\begin{align*}
x &=9\tan \theta \\
dx&=9\sec^2 \theta \, d \theta \\
81+x^2 &=81+81\tan^2 \theta \\
&=81 \sec^2 \theta.
\end{align*}
$$
We substitute,
$$
\begin{align*}
\int\frac{8}{81+x^2} \, dx &=8 \int \frac{9\sec^2 \theta \, d \theta}{81\sec^2 \theta} \\
&=\frac{8}{9}\int 1 \, d \theta \\
&=\frac{8}{9} \theta +c.
\end{align*}
$$
For the back substitution,
$$x=9\tan \theta \Rightarrow \tan \theta=\frac{x}{9} \Rightarrow \theta=\tan^{-1}\left( \frac{x}{9} \right)$$
Hence our integral,
$$\frac{8}{9}\tan^{-1}\left( \frac{x}{9} \right)+c.$$
