JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} d t=\frac{1}{2}(g(t))^2+C

, where

\mathrm{C}

is a constant, then

g(b)

is equal to :

View answer and explanation

Correct option: D

\log (2+\sqrt{5})

Option D (

\log (2+\sqrt{5})

) is correct. By substituting

u = \log(t + \sqrt{1+t^2})

, the integral simplifies, and g(t) can be found.

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