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How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$

I tried substituting $x^2+1$ as t, but it's not working

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    $\begingroup$ To be honest that is the approach to use..it would be better to show where you got lost. $\endgroup$ – Chinny84 Jul 29 '15 at 14:12
  • $\begingroup$ It's OK I got my mistake :) @Chinny84 $\endgroup$ – user220382 Jul 29 '15 at 14:18
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    $\begingroup$ Use $x^3=x(x^2+1)-x$. $\endgroup$ – Yves Daoust Jul 29 '15 at 14:42
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$$\int \frac{x^3}{(x^2+1)^{3/2}}dx=\int \frac{x(x^2+1)-x}{(x^2+1)^{3/2}}dx=\int \frac{x}{(x^2+1)^{1/2}}dx-\int \frac{x}{(x^2+1)^{3/2}}dx\\ =\sqrt{x^2+1}+\frac1{\sqrt{x^2+1}}+C$$

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$$\int\frac{x^3}{(x^2+1)^{3/2}}dx$$

$u:=x^2,du=2xdx$

$$=\frac{1}{2}\int\frac{u}{(u+1)^{3/2}}du$$

$s:=u+1,ds=du$

$$=\frac{1}{2}\int\frac{s-1}{s^{3/2}}ds$$

$$=\sqrt s+\frac{1}{\sqrt s}$$

$s=u+1,u=x^2$

$$=\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}+C$$

$$\boxed{\color{blue}{=\frac{x^2+2}{\sqrt{x^2+1}}+C}}$$

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Alternative approach:

Let $x=\tan\theta$, $dx=\sec^2\theta d\theta$

\begin{align} \int\frac{x^3dx}{(x^2+1)^{3/2}}&=\int\frac{\tan^3\theta\cdot\sec^2\theta d\theta}{\sec^3\theta}\\&=\int\frac{\sin^3\theta d\theta}{\cos^2\theta}\\&=-\int\frac{\sin^2\theta d\cos\theta}{\cos^2\theta}\\&=\int1-\sec^2\theta d\cos\theta\\&=\cos\theta+\sec\theta+C\\&=\frac1{\sqrt{1+x^2}}+\sqrt{1+x^2}+C \end{align}

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$x^2+1=t^2\Rightarrow x\,dx=t\,dt\;,\;x^2=t^2-1$ $$\int\frac{x^3}{(x^2+1)^{\frac{3}{2}}}dx=\int\frac{t^2-1}{t^2}dt=\int(1-t^{-2})dt=t+\frac{1}{t}+C$$ $t=\sqrt{1+x^2}\Rightarrow$ answer$=\sqrt{1+x^2}+\frac{1}{\sqrt{1+x^2}}+C=\frac{x^2+2}{\sqrt{x^2+1}}+C$

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Let $x^2+1=t\implies 2xdx=dt$ or $xdx=\frac{dt}{2}$ $$\int \frac{x^3 dx}{(x^2+1)^{3/2}}=\frac{1}{2}\int \frac{(t-1)dt}{(t)^{3/2}}$$ $$=\frac{1}{2}\int \frac{(t-1)dt}{t^{3/2}}$$ $$=\frac{1}{2}\int (t^{-1/2}-t^{-3/2})dt$$ $$=\frac{1}{2}\left(2t^{1/2}+2t^{-1/2}\right)$$ $$=\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}+C$$ $$=\frac{x^2+2}{\sqrt{x^2+1}}+C$$

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