The value of the integral $\smallint \,\,\frac{{1 + {x^2}}}{{1 + {x^4}}}\,\,dx$ is equal to

${\tan ^{ - 1}}\,\,{x^2}\, + \,\,C$

$\frac{1}{{2\,\sqrt 2 }}\,\,\log \,\,\left( {\frac{{{x^2}\,\, + \sqrt {\,\,2} x\,\, + \,\,1}}{{{x^2}\,\, - \,\,\sqrt 2 x\,\, + \,\,1}}} \right) + C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The value of the integral

\int \,\,\frac{{1 + {x^2}}}{{1 + {x^4}}}\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 2 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{{{x^2} - \,\,1}}{{\sqrt 2 x}}} \right)

Option B (

\frac{1}{{\sqrt 2 }}\\,{\tan ^{ - 1}}\left( {\frac{{{{{x^2}} - \\,1}}}{{\sqrt 2 x}}} \right)

) is correct because the integral is solved by dividing numerator and denominator by

x^2

and then using a suitable substitution.