The value of the
integral $\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx}$ is :
Solution
Let $I = \int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx}$<br><br>$\because$ $\log \left( {x + \sqrt {{x^2} + 1} } \right)$ is an odd function<br><br>$\therefore$ I = 0
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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