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$$\int \frac{dx}{(9+x^2)^2}$$

$x=3\tan\theta$

$dx=\frac{3}{\cos^2\theta}d\theta$

$$\int\frac{\frac{3}{\cos^2\theta}}{(9[1+\tan^2\theta])^2} \,d\theta = \int\frac{\frac{3}{\cos^2\theta}}{(\frac{9}{\cos^2\theta})^2} \, d\theta =\frac{3}{81} \int \cos^2\theta \, d\theta=\frac{3}{162}\int (\cos2\theta+1) \, d\theta$$

$u=2\theta$

$du=2d\theta$

$$\frac{3}{324}\int (\cos u+1) du=\frac{3}{324}(\sin u+u)+c=\frac{3}{324}(\sin2\theta+2\theta)+c$$

I know that $\theta=\arctan(\frac{x}{3})$

How should I continue?

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    $\begingroup$ Make a triangle that involves the arctangent by stating $tan\theta$ and derive the sine and cosine from it. Use double angle formula for $sin2\theta$ to get that in terms of sine and cosine $\endgroup$ – imranfat Feb 8 '16 at 17:55
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Setting $\tan\theta=\frac x3$ & $\theta=\tan^{-1}\left(\frac x3\right)$, one should get $$\frac{3}{324}\left(\sin 2\theta+2\theta\right)+C$$ $$=\frac{1}{108}\left(\frac{2\tan\theta}{1+\tan^2\theta}+2\theta\right)+C$$ $$=\frac{1}{108}\left(\frac{2\cdot \frac{x}{3}}{1+\frac{x^2}{9}}+2\tan^{-1}\left(\frac x3\right)\right)+C$$ $$=\frac{1}{108}\left(\frac{6x}{x^2+9}+2\tan^{-1}\left(\frac x3\right)\right)+C$$ taking $2$ common out of bracket, $$=\frac{2}{108}\left(\frac{3x}{x^2+9}+\tan^{-1}\left(\frac x3\right)\right)+C$$ $$=\color{red}{\frac{1}{54}\left(\frac{3x}{x^2+9}+\tan^{-1}\left(\frac x3\right)\right)+C}$$

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Hint. You may use $$ \sin (2\theta)=\frac{2\tan \theta}{1+\tan^2\theta} $$ and $$ \theta= \arctan \frac{x}3. $$

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Here it is: \begin{align*} \int\frac{dx}{(9+x^2)^2}\,dx &=\frac{1}{81}\int\frac{dx}{(1+(x/9)^2)^2} \;\;\;\;\;\;\;\;\mbox{put now}\;x=9\tan y \\ &=\frac{1}{81}\int\frac{1}{(1+\tan^2y)^2}\frac{9}{\cos^2y}\,dy\\ &=\frac19\int\frac{1}{\frac{1}{\cos^4y}}\frac{1}{\cos^2y}\,dy\\ &=\frac19\int{\cos^2y}\,dy\\ &=\frac1{18}\left[\cos y\sin y +y\right]+C\\ &=\frac1{18}\left[\cos^2 y\tan y +y\right]+C\\ \end{align*} but now $y=\arctan x/9$, from which the last expression turns into $$ \frac{x\cos^2(\arctan(x/9))}{18\cdot9}+\frac{\arctan(x/9)}{18}+C $$

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