JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

\int {\frac{{{x^2} + {{\sin }^2}x}}{{1 + {x^2}}}{{\sec }^2}xdx}

is equal to

View answer and explanation

Correct option: B

\tan x - {\tan ^{ - 1}}x + c

Option B (

\frac{\left(x-\frac{1}{x}\right)^{K+8}}{k+8}+c

) is correct because after simplifying and substituting

t = x - \frac{1}{x}

, the integral becomes a simple power rule integral.

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