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Find $\int \frac{x}{\sqrt{x^{2}+4}} d x$

$\sqrt{x^{2}+a^{2}}$

$x=2 \tan \theta$

let $u=x^{2}+4$

$d u=2 x d x$

$\int \frac{x}{\sqrt{u}} d x=\int \frac{x}{\sqrt{u}} \frac{2}{2} d x$

$=\frac{1}{2} \int \frac{2 x}{\sqrt{u}} d x=\frac{1}{2} \int \frac{d u}{\sqrt{u}}$

$=\frac{1}{2} \int \frac{d u}{u^{\frac{1}{2}}}=\frac{1}{2} \int u^{-\frac{1}{2}} d u=\frac{1}{2}\left[2 u^{\frac{1}{2}}\right]+c$

$=u^{\frac{1}{2}}+c$

$\int \frac{x}{\sqrt{x^{2}+4}} d x=u^{\frac{1}{2}}+c$

$=\left(x^{2}+u\right)^{\frac{1}{2}}+c=\sqrt{x^{2}+u}+c$

Find $\int_{0}^{\frac{3 \sqrt{3}}{2}} \frac{x^{3}}{\left(4 x^{2}+9\right)^{3 / 2}} d x$

$\left(4 x^{2}+9\right)^{3 / 2}=\sqrt{\left(4 x^{2}+9\right)}^{3}$

$\sqrt{x^{2}+a^{2}} \rightarrow x=a \tan \theta$

let $u=2 x \rightarrow d{u}=2 d x \rightarrow d x=\frac{d u}{2}$

$u=2 x \rightarrow x=\frac{u}{2}$

when $x=0 \quad \rightarrow u=0$

when $x=\frac{3 \sqrt{3}}{2} \rightarrow u=3 \sqrt{3}$

$u=2 x \rightarrow u^{2}=u x^{2}$

$I=\int_{0}^{\frac{3 \sqrt{3}}{2}} \frac{x^{3}}{\left(\sqrt{4 x^{2}+9}\right)^{3}} \cdot d x=\int_{0}^{3 \sqrt{3}} \frac{\left(\frac{u}{2}\right)^{3}}{\left(\sqrt{u^{2}+9}\right)^{3}} \cdot d x$

$=\int_{0}^{3 \sqrt{3}} \frac{\frac{u^{3}}{8}}{\left(\sqrt{u^{2}+9}\right)^{3}} \cdot \frac{d u}{2}$

$I=\frac{1}{16} \int_{0}^{3 \sqrt{3}} \frac{u^{3}}{(\sqrt {u^{2}+9})^{3}} \cdot d u$

let $u=3 \tan \theta \rightarrow d u=3 \sec ^{2} \theta d \theta, \quad \frac{-\pi}{2} \le \theta \le \frac{\pi}{2}$

$\sqrt{u^{2}+9}=3 \sec \theta$

when $u={0} \rightarrow \tan \theta =0 \rightarrow \theta =\tan ^{-1}(0)=0$

when $u=3 \sqrt{3} \rightarrow \tan \theta=\frac{u}{3}=\frac{3 \sqrt{3}}{3}=\sqrt{3} \:, \theta = \tan^{-1} ({\sqrt 3})$

$=\frac{\pi}{3}$

$I = {\frac{1}{16}} \int_{0}^{3\sqrt 3} \frac{u^{3}}{\left(\sqrt{u^{2}+9}\right)^{3}} \cdot d u$

$0, \frac{\pi}{3}, \quad u=3 \tan \theta, \sqrt{u^{2}+9}=3 \sec \theta$

$d u=3 \sec ^{2} \theta d \theta$

$I=\frac{1}{16} \int_{0}^{\frac{\pi}{3}} \frac{(3 \tan \theta)^{3}}{(3 \sec \theta)^{3}} \cdot 3 \sec ^{2} \theta d \theta$

$=\frac{1}{16} \int_{0}^{\frac{\pi}{3}} \frac{9 \tan ^{3} \theta}{9 \sec ^{3} \theta} \cdot 3 \sec ^{2} \theta d \theta$

$I=\frac{3}{16} \int_{0}^{\frac{\pi}{3}} \frac{\tan ^{3} \theta}{\sec ^{3} \theta} \cdot \sec ^{2} \theta d \theta$

$=\frac{3}{16} \int_{0}^{\frac{\pi}{3}} \frac{\tan ^{3} \theta}{\sec \theta} \cdot d \theta$

$= \frac{3}{16} \int_{0}^{\frac{\pi}{3}} \frac{\tan ^{2} \theta}{\sec \theta} \cdot \tan \theta \cdot \frac{\sec \theta}{\sec \theta} \cdot d \theta$

but I have $\tan ^{2} \theta=\sec ^{2} \theta-1$

$I=\frac{3}{16} \int_{0}^{\frac{\pi}{3}} \frac{\left(\sec ^{2} \theta-1\right)}{\sec ^{2} \theta} \cdot \tan \theta \cdot \sec \theta \cdot d \theta$

let $t=\sec \theta \rightarrow d t=\sec \theta \tan \theta d \theta$

when $\theta=0 \quad \rightarrow t=\sec (0)=1$

when $\theta=\frac{\pi}{3} \rightarrow t=\sec \left(\frac{\pi}{3}\right)=2$

$I=\frac{3}{16} \int_{1}^{2} \frac{\left(t^{2}-1\right)}{t^{2}} \cdot d t$

$=\frac{3}{16} \int_{1}^{2}\left(\frac{t^{2}}{t^{2}}-\frac{1}{t^{2}}\right) d t$

$=\frac{3}{16} \int_{1}^{2} 1-t^{-2} d t=\frac{3}{16}\left[t+\frac{1}{t}\right]_{1}^{2}$

$=\frac{3}{16}\left[2+\frac{1}{2}-\left(1+\frac{1}{1}\right)\right]$

$=\frac{3}{16}\left[2+\frac{1}{2}-1-1\right]=\frac{3}{16}\left[\frac{1}{2}\right]=\frac{3}{32}$

$I = \int_{0}^{\frac{3 \sqrt{3}}{2}} \frac{x^{3}}{\left(\sqrt{\left(4 x^{2}+9\right)}\right)^{3 / 2}} d x=\frac{3}{32}$