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.
 A: First get $\frac{8}{81}$ out of the integral (linearity) and you are left with: $\int\frac{1}{1+ (x/9)^2}dx$, now transform: $t = x/9$ so $9dt= dx$. You obtain 
$\frac{8}{81}\int\frac{9}{1+t^2}dt$. And now you of course obtain 
$$
\begin{align}
\int\frac{8}{81+x^2}dx &=\frac{8}{9} \arctan (t) + c \\
                       &= \frac{8}{9} \arctan (x/9) +c
\end{align}
$$
A: Hint: $$\int \frac{8}{81+x^2}\mathop{dx} = \frac{8}{81}\int\frac{1}{1+(x/9)^2}\mathop{dx}=\frac{8}{81}\int\frac{1}{1+u^2}9\mathop{du}$$
A: Hint: Try to transform the integrand into something resembling $\frac{1}{1+\text{something}^2}$. Use the substitution $u = \text{something}$.
A: 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,
$$\frac{8}{9}\tan^{-1}\left( \frac{x}{9} \right)+c.$$
