Question

Solve the integral.

$\int \frac{x^2}{\sqrt{x^2+1}}dx$

Short answer

The solution is $\frac{x}{2}·\sqrt{x^2+1}-\frac{1}{2}\sin h^{-1}x+C$.

Step-by-step solution

Step 1. Given information.

The given integral is$\int \frac{x^2}{\sqrt{x^2+1}}dx$.

Step 2. Solve the integral.

Let $I=\int \frac{x^2}{\sqrt{x^2+1}}dx$.

Substitute $x=\sin hy$, $dx=\cos hy·dy$.

$\begin{array}{rcl}I& =& \int \frac{{\left(\sin hy\right)}^2·\cos hy}{\sqrt{{\left(\sin hy\right)}^2+1}}dy\\ & =& \int \frac{{\left(\sin hy\right)}^2·\cos hy}{\sqrt{{\left(\cos hy\right)}^2}}dy\\ & =& \int \frac{{\left(\sin hy\right)}^2·\cos hy}{\cos hy}dy\\ & =& \int {\left(\sin h\right)}^{2}dy\\ & =& \int \frac{\cos h2y-1}{2}dy\\ & =& \frac{1}{4}\sin h2y-\frac{y}{2}+C\\ & =& \frac{1}{4}·2\sin hy·\cos hy-\frac{y}{2}+C\\ & =& \frac{1}{2}\sin hy·\cos hy-\frac{y}{2}+C\end{array}$

Step 3. Simplified answer.

$I=\frac{1}{2}\sin hy·\cos hy-\frac{y}{2}+C$

Substitute $x=\sin hy,\cos hy=\sqrt{x^2+1}$and $y=\sin h^{-1}x$.

So, the required solution is:

$I=\frac{x}{2}·\sqrt{x^2+1}-\frac{1}{2}\sin h^{-1}x+C$