Mathematics

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


ANSWER

$${1 \over 2}\log \left( {\left( {{x^2} + {1 \over 2}} \right) + \sqrt {{x^4} + {x^2} + 1} } \right) + c$$


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

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

then,
$$\Rightarrow I=\dfrac12\int\dfrac{1}{\sqrt{t^2 +t+1}}dt$$

$$\Rightarrow I=\dfrac12\int\dfrac{1}{\sqrt{t^2 +t+\dfrac14-\dfrac14+1}}dt$$

$$\Rightarrow I=\dfrac12\int\dfrac{1}{\sqrt{(t+\dfrac12)^2+ (\dfrac{\sqrt3}{2})^2}}dt$$

$$\Rightarrow \dfrac12log\begin{pmatrix}(t+\dfrac12)+\sqrt{(t+\dfrac12)^2+(\dfrac{\sqrt3}{2})^2}\end{pmatrix}$$

from
$$\begin{bmatrix}\because \int\dfrac{1}{\sqrt{a^2+x^2}}dx = log\begin{bmatrix} x+ \sqrt{x^2+a^2}\end{bmatrix}+ C\end{bmatrix}$$

 again put $$t=  x^2$$

$$\Rightarrow \dfrac12log\begin{pmatrix}(x^2+\dfrac12)+\sqrt{(x^2+\dfrac12)^2+(\dfrac{\sqrt3}{2})^2}\end{pmatrix} + C$$


$$= \dfrac12log\begin{pmatrix}(x^2+\dfrac12)+\sqrt{(x^4+x^2+1}\end{pmatrix} + C$$

Hence Correct answer is $$B$$.
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Single Correct Medium Published on 17th 09, 2020
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