Question

Evaluate ${\int }_0^1\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{3}$

$l={\int }_0^1\frac{1-x}{1+x}\cdot \frac{dx}{\sqrt{x+x^2+x^3}}$
$={\int }_0^1\frac{(1-x^2)}{(x^2+2x+1)\sqrt{x+x^2+x^3}}$
$={\int }_0^1\frac{\frac{1}{x^2}-1dx}{(x+\frac{1}{x}+2)\sqrt{x+\frac{1}{x}+1}}$
Put $x+\frac{1}{x}+1=t^2\Rightarrow (1-\frac{1}{x^2}dx)=2tdt$
$l={\int }_{\infty }^{\sqrt{3}}\frac{-2tdt}{(t^2+1)t}=2{\int }_{\sqrt{3}}^{\infty }\frac{dt}{t^2+1}=2{\left[{\tan }^{-1}\right]}_{\sqrt{3}}^{\infty }=\frac{\pi }{3}$