How to integrate $\frac{x^2}{x^2+9}$? I know you can use division to get $1-\frac{p}{x^2+9}$ and then use the result for arctan, but I was trying to do this using substitution and nothing seems to work? Is there a substitution that could be used for this?
 A: If one chooses not to write $\frac{x^2}{x^2+9}=1-\frac{9}{x^2+9}$, then one can make still directly substitute $x=3 \tan y$ with $dx=3 \sec^2y \,dy$.  Thus, we have
$$\begin{align}
\int \frac{x^2}{x^2+9}dx&=\int\frac{9\tan^2y}{9\tan^2y+9}3\sec^2y\,dy\\\\
&=3\int \tan^2y\,dy\\\\
&=3(\tan y-y)+C\\\\
&=x-3\arctan (x/3)+C
\end{align}$$
A: $\frac{9}{x^{2}+9}=\frac{9}{9((\frac{x}{3})^{2}+1)}$ so how about $u=\frac{x}{3}$? and then you can use the fact that $\frac{x^{2}}{x^{2}+9}=1-\frac{9}{x^{2}+9}$, and the antiderivative will be $x -3\arctan(\frac{x}{3})+C$, where $C$ is a constant.
A: You can try to do $x = iz, dx = idz$, will by conjugate rule give $\frac{-z^2idz}{(3+z)(3-z)}$, then try and then proceed by some method. This may even give insight as to the popular complex extension of arctan using complex logarithms in one complex variable.
A: Let $x=3\tan\theta$, $dx=3\sec^2\theta d\theta$.
\begin{align}
\int\frac{x^2 dx}{x^2+9}&=\int1-\frac9{x^2+9}dx\\&=x-\int\frac{27\sec^2\theta d\theta}{9\sec^2\theta}\\&=x-3\int d\theta \\&=x-3\tan^{-1}{\frac x 3}+C 
\end{align} 
In case you don't understand the last step,
$$\frac x 3=\tan\theta\implies\theta=\tan^{-1}{\frac x3}$$
