Question

$\int \frac{x^{3}dx}{\sqrt{1+x^2}}$ is equal to

• A. $\frac{1}{3}\sqrt{1+x^2}(2+x^2)+c$
• B. $\frac{1}{3}\sqrt{1+x^2}(x^2-1)+c$
• C. $\frac{1}{3}(1+x^2{)}^{3/2}+c$
• D. $\frac{1}{3}\sqrt{1+x^2}(x^2-2)+c$

Answer 1

The correct option is D $\frac{1}{3}\sqrt{1+x^2}(x^2-2)+c$

Let $I=\int \frac{x^{3}dx}{\sqrt{1+x^2}}$

Substitute $t=x^2\Rightarrow dt=2xdx$

$I=\frac{1}{2}\int \frac{t}{\sqrt{t+1}}dt$

Substitute $u=t+1\Rightarrow du=dt$
$I=\frac{1}{2}\int \frac{u-1}{\sqrt{u}}du=\frac{1}{2}\int (\sqrt{u}-\frac{1}{\sqrt{u}})du$
$=\frac{u^{3/2}}{3}-\sqrt{u}=\frac{1}{3}(x^2-2)\sqrt{1+x^2}$