Question

Evaluate: $\int \frac{dx}{x\sqrt{x^2+x+1}}$

• A. $ln\left|\frac{\sqrt{x^2-x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|+c$
• B. $ln\left|\frac{\sqrt{x^2-x+1}+x-1}{\sqrt{x^2-x+1}+x+1}\right|+c$
• C. $ln\left|\frac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|+c$
• D. None of these

Answer 1

Answer: (C)

$ln\left|\frac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|+c$

Given :-

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

To evaluate the integral

Sol.:

$\sqrt{x^{2}x+1}=\pm \left(x\right)+t\left(UsingEule{r}^{\prime }sSubstitution\right)$

$x^2+x+1=x^2+2xt+t^2$

$x=\frac{1-t^2}{2t-1}$

$dx=\frac{-2(t^2-t+1)}{2t-1}dt$

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

$I=\int (\frac{\frac{-2(t^2-t+1)}{2t-1}}{\frac{(1-t^2)(t^2-t+1)}{(2t-1{)}^2}}dt$

$I=\int (\frac{-2(2t-1)}{(1-t^2)}dt$

$I=2(\int (\frac{(2t-1)}{(t^2-1)}dt)$

$I=2(\int \frac{2t}{(t^2-1)}dt)-\int \left(\frac{2}{(t^2-1)}\right)dt$

$I=ln\left(\left|\right(t^2-1\left)\right|\right)-ln\left(\left|\right(\frac{t-1}{t+1}\left)\right|\right)+C$

$I=2ln\left(\left|\right(t+1\left)\right|\right)+2ln\left(\left|\right(t-1\left)\right|\right)-ln\left(\left|\right(t-1\left)\right|\right)-ln\left(\left|\right(t+1\left)\right|\right)+C$

Hence the correct is

$ln\left|\frac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|+C$