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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?

David Martínez
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Meep
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    $\frac{x^2}{x^2+9}=\frac{x^2+9-9}{x^2+9}=1-\frac{9}{x^2+9}$ could help. Now, a clear substitution would take you to $\tan^{-1}(.)$. – Claude Leibovici Jun 21 '15 at 14:39

4 Answers4

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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

$$\begin{align} \int \frac{x^2}{x^2+9}dx&=\int\frac{9\tan^2y}{9\tan^2y+9}3\sec^2y\,dy\\\\ &=3\int \tan^2y\,dy\\\\ &=3(\tan y-y)+C\\\\ &=x-3\arctan (x/3)+C \end{align}$$

Mark Viola
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$\frac{9}{x^{2}+9}=\frac{9}{9((\frac{x}{3})^{2}+1)}$ 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.

mich95
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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.

mathreadler
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    Oh look, someone already did this: http://math.stackexchange.com/questions/1102389/arctan-in-complex-logarithm-form – mathreadler Jun 21 '15 at 15:26
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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,

$$\frac x 3=\tan\theta\implies\theta=\tan^{-1}{\frac x3}$$

Mythomorphic
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