JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int\limits_{0}^1 \frac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}} \mathrm{~d} x=\log \frac{e+A}{1+\sqrt{2}}

then

A

is equal to

View answer and explanation

Correct option: B

\sqrt{e^2+1}

Option B (

\sqrt{{e^2+1}}

) is correct because after substituting

e^x = t

, the integral evaluates to

\log \frac{{e+\sqrt{{e^2+1}}}}{{1+\sqrt{{2}}}}

, thus

A = \sqrt{{e^2+1}}

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