Question

The value of $\int \frac{x}{\sqrt{x^4+x^2+1}}dx$ equals

• A. $\frac{1}{2}\log \left(x^2+\sqrt{x^4+x^2+1}\right)+c$
• B. $\frac{1}{2}\log \left(\left(x^2+\frac{1}{2}\right)+\sqrt{x^4+x^2+1}\right)+c$
• C. $\log \left(\left(x^2+\frac{1}{2}\right)+\sqrt{x^4+x^2+1}\right)+c$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}\log \left(\left(x^2+\frac{1}{2}\right)+\sqrt{x^4+x^2+1}\right)+c$

we have to evaluate $I=\int \frac{x}{\sqrt{x^4+x^2+1}}dx$

Let us assume $x^2=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}$

then,

$\Rightarrow I=\frac{1}{2}\int \frac{1}{\sqrt{t^2+t+1}}dt$

$\Rightarrow I=\frac{1}{2}\int \frac{1}{\sqrt{t^2+t+\frac{1}{4}-\frac{1}{4}+1}}dt$

$\Rightarrow I=\frac{1}{2}\int \frac{1}{\sqrt{(t+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}}dt$

$\Rightarrow \frac{1}{2}log\left(\begin{array}{c}(t+\frac{1}{2})+\sqrt{(t+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}\end{array}\right)$

from

$\left[\begin{array}{c}\because \int \frac{1}{\sqrt{a^2+x^2}}dx=log\left[\begin{array}{c}x+\sqrt{x^2+a^2}\end{array}\right]+C\end{array}\right]$

again put $t=x^2$

$\Rightarrow \frac{1}{2}log\left(\begin{array}{c}(x^2+\frac{1}{2})+\sqrt{(x^2+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}\end{array}\right)+C$

$=\frac{1}{2}log\left(\begin{array}{c}(x^2+\frac{1}{2})+\sqrt{(x^4+x^2+1}\end{array}\right)+C$

Hence Correct answer is $B$.