Question

If $\int \frac{x^2-1}{{\left(x+1\right)}^2\sqrt{x\left(x^2+x+1\right)}}dx=A{\tan }^{-1}\frac{\sqrt{x^2+x+1}}{x}+c$ where $c$ is real constant, then $A=$

• A. $\frac{1}{2}$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C

$2$

Explanation for the correct option.

$\begin{array}{rcl}\frac{d}{dx}\left({\tan }^{-1}\frac{\sqrt{x^2+x+1}}{x}\right)& =& \frac{1}{1+{\displaystyle \frac{x^2+x+1}{x}}}·\frac{1}{2}\sqrt{\frac{x}{x^2+x+1}}\left\{\frac{x\left(2x+1\right)-\left(x^2+x+1\right)}{x^2}\right\}\\ & =& \frac{x}{2{\left(x+1\right)}^2}\sqrt{\frac{x}{x^2+x+1}}\left\{\frac{x^2-1}{x^2}\right\}\\ & =& \frac{x-1}{2x\left(x+1\right)}\sqrt{\frac{x}{x^2+x+1}}\\ & =& \frac{x^2-1}{2{\left(x+1\right)}^2\sqrt{x\left(x^2+x+1\right)}}\end{array}$

Substituting this in the given equation, we get

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

Therefore, $A=2$

Hence, option C is correct.