Finding an Integral Containing \(\sqrt{x^{2}+9}\)

Find \(\int \frac{dx}{(x^{2}+9)^{3/2}}\).

Solution The integral contains a square root \((x^{2}+9)^{3/2}= \big( \sqrt{x^{2}+9}\big) ^{3}\) that is of the form \(\sqrt{x^{2}+a^{2}},\) where \(a=3\). We use the substitution \(x=3\tan \theta\), \(-\dfrac{\pi }{2} \lt \theta \lt \dfrac{\pi }{2}\). Then \(dx=3\sec ^{2}\theta d\theta\). Since \[ \begin{eqnarray*} (x^{2}+9)^{3/2}\underset{\underset{\color{#0066A7}{\hbox{\(x=3\tan \theta\)}}}{\color{#0066A7}{\uparrow}}}{=}(9\tan^{2}\theta +9)^{3/2}=9^{3/2}(\tan^{2}\theta +1)^{3/2}\underset{\underset{\color{#0066A7}{\hbox{\(\tan^{2}\theta +1=\sec ^{2}\theta\)}}}{\color{#0066A7}{\uparrow}}}{=}27(\sec^{2}\theta)^{3/2}\underset{\underset{\color{#0066A7}{\hbox{\(\sec \theta >0\)}}}{\color{#0066A7}{\uparrow}}}{=}27\sec ^{3}\theta \end{eqnarray*} \]

we have \[ \begin{equation*} \int \frac{dx}{(x^{2}+9)^{3/2}}=\int \frac{3\sec ^{2}\theta d\theta }{27\sec ^{3}\theta }=\frac{1}{9}\int \frac{d\theta }{ \sec \theta }=\frac{1}{9}\int \cos \theta d\theta =\frac{1}{9}\sin \theta +C \end{equation*} \]

To express the solution in terms of \(x,\) use either the right triangles in Figure 5 or identities.

From the right triangles, the hypotenuse is \(\sqrt{x^{2}+3^{2}}=\sqrt{x^{2}+9 }\). So, \(\sin \theta =\dfrac{x}{\sqrt{x^{2}+9}}.\) Then \[ \int \frac{dx}{(x^{2}+9)^{3/2}}=\dfrac{1}{9}\sin \theta +C=\dfrac{x}{9 \sqrt{x^{2}+9}}+C \]