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$$\int x\sqrt{x+x^2}\,dx = ?$$

Can you solve this problem? I have tried this after a challenge. ðŸ˜Š

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**Hint:**

Set $x^2+x=u$, we have

$$\int x\sqrt{x+x^2}\,dx =\frac 12 \int (2x+1)\sqrt{x+x^2}\,dx-\frac 12 \int \sqrt{-\frac 14+\left(x+\frac 12\right)^2}\,dx$$

Let $x+\frac12=\frac12\cosh(u)$, then $\sqrt{x+x^2}=\frac12\sinh(u)$ and $\mathrm{d}x=\frac12\sinh(u)\,\mathrm{d}u$. Furthermore, $\sinh^2(u)=\frac{\cosh(2u)-1}2$

$$

\begin{align}

\int x\sqrt{x+x^2}\,\mathrm{d}x

&=\frac18\int(\cosh(u)-1)\sinh^2(u)\,\mathrm{d}u\\

&=\frac1{24}\sinh^3(u)-\frac1{32}\sinh(2u)+\frac1{16}u+C\\

&=\frac1{24}\sinh^3(u)-\frac1{16}\sinh(u)\cosh(u)+\frac1{16}u+C\\

&=\frac13\sqrt{x+x^2}^{\,3}-\frac18(2x+1)\sqrt{x+x^2}+\frac1{16}\cosh^{-1}(2x+1)+C

\end{align}

$$

Hint: You can do that easily using Euler’s substitution.

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