Question

$\int \frac{1}{(x+1)\sqrt{x^2+x+1}}dx$

• A. $=-\log |\frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{x^2+x+1}}{x+1}|+C$
• B. $=-\log |\frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{x^2+x+1}}{x-1}|$
• C. $=-\log |\frac{1}{x-1}-\frac{1}{2}+\frac{\sqrt{x^2+x-1}}{x-1}|+C$
• D. $=-\log |\frac{1}{x-1}-\frac{1}{2}+\frac{\sqrt{x^2+x+1}}{x+1}|+C$

Answer 1

The correct option is A $=-\log |\frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{x^2+x+1}}{x+1}|+C$

$I=\int \frac{1}{(x+1)\sqrt{x^2+x+1}}dx$
Let, $x+1=\frac{1}{t}$ or $dx=-\frac{1}{t^2}dt$
$\therefore I=\int \frac{1}{\frac{1}{t}\sqrt{{(\frac{1}{t}-1)}^2+(\frac{1}{t}-1)+1}}(-\frac{1}{t^2})dt$
$=-\int \frac{dt}{\sqrt{{(1-t)}^2+(t-t^2)+t^2}}$
$=-\int \frac{dt}{\sqrt{t^2-t+1}}=-\int \frac{dt}{\sqrt{{(t-\frac{1}{2})}^2+\frac{3}{4}}}$
$=-\log |(t-\frac{1}{2})+\sqrt{t^2-t+1}|+C$
$=-\log |\frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{x^2+x+1}}{x+1}|+C$