Question

Evaluate $\int \sqrt{x+\sqrt{x^2+2}}dx$.

• A. $\frac{1}{3}{(x+\sqrt{x^2+2})}^{\frac{3}{2}}-2\frac{1}{\sqrt{x+\sqrt{x^2+2}}}+C$
• B. $\frac{1}{3}{(x+\sqrt{x^2+2})}^{\frac{1}{2}}+2\frac{1}{\sqrt{x+\sqrt{x^2+2}}}+C$
• C. $\frac{2}{3}{(x+\sqrt{x^2+2})}^{\frac{2}{3}}+2\frac{1}{\sqrt{x+\sqrt{x^2+2}}}+C$
• D. None of these

Answer 1

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

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

Let $x+\sqrt{x^2+2}=p$

$\Rightarrow x^2+2=p^2+x^2-2px$

$\Rightarrow x=\frac{p}{2}-\frac{1}{p}$

$\Rightarrow dx=(\frac{1}{2}+\frac{1}{p^2})dp=\frac{p^2+2}{2p^2}dp$
$I=\int \frac{p^{\frac{1}{2}}(p^2+2)}{2p^2}dp$
$I=\frac{1}{2}\int p^{\frac{1}{2}}dp+\int p^{\frac{-3}{2}}dp$

$=\frac{1}{3}{\left(p\right)}^{\frac{3}{2}}-2\frac{1}{\sqrt{p}}+C$
$=\frac{1}{3}{(x+\sqrt{x^2+2})}^{\frac{3}{2}}-2\frac{1}{\sqrt{x+\sqrt{x^2+2}}}+C$