Example 3.28.
Let us compute \(\ds \int \frac{dx}{(x^2+9)^2}\text{.}\) The discussion above suggests the substitution \(x = 3\tan \theta\text{.}\) So, \(dx =
3\sec^2(\theta)d\theta\) and \(x^2 + 9 = 9\sec^2(\theta)\text{.}\) Thus,
\begin{align*}
\int \frac{dx}{(x^2+9)^2} &= \int
\frac{3\sec^2\theta}{(9\sec^2\theta)^2} d\theta = \frac{1}{27}\int
\cos^2(\theta)d\theta.
\end{align*}
The last integral can be handled by techniques in SectionΒ 3.2
\begin{align*}
\int \cos^2(\theta)d\theta &= \int \frac{\cos(2\theta) +1}{2}
d\theta \\
&= \frac{1}{2}\left( \frac{1}{2}\sin(2\theta)
+ \theta \right) +C\\
&= \frac{1}{2}\left(
\sin\theta\cos\theta + \theta \right) + C.
\end{align*}
From the corresponding diagram (FigureΒ 3.27 with \(a=3\)) we get \(\theta = \arctan(x/3)\) and that
\begin{equation*}
\sin(\theta)\cos(\theta) = \frac{x}{\sqrt{x^2+9}}\frac{3}{\sqrt{x^2+9}}
= \frac{3x}{x^2+9}.
\end{equation*}
Hence,
\begin{equation*}
\int \frac{dx}{(x^2+9)^2} = \frac{1}{54}\left(\frac{3x}{x^2+9} +
\arctan\left( \frac{x}{3} \right)\right) + C.
\end{equation*}