Use $u$ substitution:

$u = 1 + x^2$, $du = 2xdx$ , $dx =\frac {du}{2x}$

Write the given integral in terms of : $u$, $du$ and $dx$

$$\frac {1}{2}\int \frac {x^4}{x\sqrt{u}} du \implies \frac{1}{2} \int \frac {u+x^2 - 1}{x \sqrt{u}}$$

This works right I double checked my work. But it looks awfully complicated for a calculus problem than the other ones I have done.

  • $\begingroup$ Other obvious substitutions are $x=\tan t$ or $x=\sinh u$. $\endgroup$ – Lucian Oct 6 '14 at 22:52

Set $\sqrt{1+x^2}=u\implies 1+x^2=u^2, xdx=u du$


  • $\begingroup$ I'm a bit confused when you wrote xdx = udu. what do you mean. $\endgroup$ – user983246 Oct 6 '14 at 4:56
  • $\begingroup$ @user983246, Differentiate either sides of $$1+x^2=u^2$$ $\endgroup$ – lab bhattacharjee Oct 6 '14 at 4:57
  • $\begingroup$ $u^2 -1$ only adds up to $x^2$ in your case $\endgroup$ – user983246 Oct 6 '14 at 5:02
  • $\begingroup$ @user983246, $$\frac{d(1+x^2)}{dx}=\frac{d(u^2)}{dx} \frac{d(1+x^2)}{dx}=\frac{d(u^2)}{du}\frac{du}{dx}$$ $$\implies2x\ dx=2u\cdot\frac{du}{dx}$$ $\endgroup$ – lab bhattacharjee Oct 6 '14 at 5:03

Put $$u=\sqrt{1+{{x}^{2}}}\Rightarrow {{u}^{2}}=1+{{x}^{2}}\Rightarrow 2udu=2xdx\Rightarrow xdx=udu$$ Then $$\begin{align} & I=\int{\frac{{{x}^{3}}}{\sqrt{1+{{x}^{2}}}}dx}=\int{\frac{{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}xdx}=\int{\frac{{{u}^{2}}-1}{u}udu}=\int{\left( {{u}^{2}}-1 \right)du} \\ & =\frac{{{u}^{3}}}{3}-u+C=\frac{1}{3}{{\left( \sqrt{1+{{x}^{2}}} \right)}^{3}}-\sqrt{1+{{x}^{2}}}+C \\ \end{align}$$


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