Question

The value of ${\int }_0^1\frac{dx}{(x+1)\sqrt{x^2+2x}}$ is

• A. $\pi /6$
• B. $\pi /3$
• C. $\pi /2$
• D. $\pi$

Answer 1

Answer: (B)

$\pi /3$

Let $I={\int }_0^1\frac{1}{(x+1)\sqrt{x^2+2x}}dx={\int }_0^1\frac{1}{(x+1)\sqrt{{(x+1)}^2-1}}dx$
Substitute $u=x+1\Rightarrow du=dx$
$\therefore I={\int }_1^2\frac{1}{u\sqrt{u^2-1}}du$
Substitute $s=\sqrt{u^2-1}\Rightarrow ds=\frac{u}{\sqrt{u^2-1}}du$
$\therefore I={\int }_0^{\sqrt{3}}\frac{1}{s^2+1}ds={\left[{\tan }^{-1}s\right]}_0^{\sqrt{3}}=\frac{\pi }{3}$