Question

# The given equation $\int {\dfrac{{1 - {x^2}}}{{(1 + {x^2})\sqrt {1 + {x^4}} }}dx}$ is equal to: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

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