JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 7)
The value of the
integral $$\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} $$ is :
integral $$\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} $$ is :
2
0
$$-$$1
1
Explanation
Let $$I = \int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} $$
$$\because$$ $$\log \left( {x + \sqrt {{x^2} + 1} } \right)$$ is an odd function
$$\therefore$$ I = 0
$$\because$$ $$\log \left( {x + \sqrt {{x^2} + 1} } \right)$$ is an odd function
$$\therefore$$ I = 0
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