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A.$\sqrt 2 {\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$

B.$\dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$

C.$\dfrac{1}{2}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$

D.None of these

Answer

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Hint: Use substitution method of integration and let's take $\dfrac{1}{x} + x$ as t and then solve with respect to t first and later substitute its value.

Consider the given integral as I. Now,

$\

I = \int {\dfrac{{1 - {x^2}}}{{(1 + {x^2})\sqrt {1 + {x^4}} }}dx} \\

\Rightarrow I = \int {\dfrac{{{x^2}(\dfrac{1}{{{x^2}}} - 1)}}{{{x^2}(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\

\Rightarrow I = \int {\dfrac{{(\dfrac{1}{{{x^2}}} - 1)}}{{(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\

\ $

Observe that, If, we consider $\dfrac{1}{x} + x$ as t then, on differentiating, we can get $ - \dfrac{1}{{{x^2}}} + 1$ as $dt$ .Using these information,

$\

\Rightarrow I = \int {\dfrac{{(\dfrac{1}{{{x^2}}} - 1)}}{{(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\

\Rightarrow I = \int {\dfrac{{ - dt}}{{t\sqrt {{t^2} - {{(\sqrt 2 )}^2}} }}{\text{ }}[{x^2} + \dfrac{1}{{{x^2}}} = {{(x + \dfrac{1}{x})}^2} - 2]} \\

\Rightarrow I = - \int {\dfrac{{dt}}{{t\sqrt {{t^2} - {{(\sqrt 2 )}^2}} }}} \\

\Rightarrow I = \dfrac{1}{{\sqrt 2 }}\cos e{c^{ - 1}}(\dfrac{{x + \dfrac{1}{x}}}{{\sqrt 2 }}) + c \\

\Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c \\

\ $

Hence the given integral $I = \dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$

Option B is correct.

Note: There is one more method to solve this integral by substitution x with a trigonometric function. We found that this method is easier to understand.

Consider the given integral as I. Now,

$\

I = \int {\dfrac{{1 - {x^2}}}{{(1 + {x^2})\sqrt {1 + {x^4}} }}dx} \\

\Rightarrow I = \int {\dfrac{{{x^2}(\dfrac{1}{{{x^2}}} - 1)}}{{{x^2}(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\

\Rightarrow I = \int {\dfrac{{(\dfrac{1}{{{x^2}}} - 1)}}{{(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\

\ $

Observe that, If, we consider $\dfrac{1}{x} + x$ as t then, on differentiating, we can get $ - \dfrac{1}{{{x^2}}} + 1$ as $dt$ .Using these information,

$\

\Rightarrow I = \int {\dfrac{{(\dfrac{1}{{{x^2}}} - 1)}}{{(\dfrac{1}{x} + x)\sqrt {\dfrac{1}{{{x^2}}} + {x^2}} }}dx} \\

\Rightarrow I = \int {\dfrac{{ - dt}}{{t\sqrt {{t^2} - {{(\sqrt 2 )}^2}} }}{\text{ }}[{x^2} + \dfrac{1}{{{x^2}}} = {{(x + \dfrac{1}{x})}^2} - 2]} \\

\Rightarrow I = - \int {\dfrac{{dt}}{{t\sqrt {{t^2} - {{(\sqrt 2 )}^2}} }}} \\

\Rightarrow I = \dfrac{1}{{\sqrt 2 }}\cos e{c^{ - 1}}(\dfrac{{x + \dfrac{1}{x}}}{{\sqrt 2 }}) + c \\

\Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c \\

\ $

Hence the given integral $I = \dfrac{1}{{\sqrt 2 }}{\sin ^{ - 1}}(\dfrac{{\sqrt 2 x}}{{{x^2} + 1}}) + c$

Option B is correct.

Note: There is one more method to solve this integral by substitution x with a trigonometric function. We found that this method is easier to understand.