The value of the integral ∫ - 1 1 log x + x 2 + 1 d x is:

Let I = ∫ - 1 1 log x + x 2 + 1 d x Consider f x = log x + x 2 + 1 Now, f - x = log - x + 1 + x 2 ⇒ f - x = - log x + 1 + x 2 ⇒ f - x = - f x So, f x is an odd function. ∴ I = 0

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties of definite integrationEasy2 minPYQ_2021

MathematicsEasysingle choice

The value of the integral $\int_{- 1}^{1} \log (x + \sqrt{x^{2} + 1}) d x$ is:

Options:

Answer:

Solution:

Let $I = \int_{- 1}^{1} \log (x + \sqrt{x^{2} + 1}) d x$

Consider $f (x) = \log (x + \sqrt{x^{2} + 1})$

Now, $f (- x) = \log (- x + \sqrt{1 + x^{2}})$

$\Rightarrow f (- x) = - \log (x + \sqrt{1 + x^{2}})$

$\Rightarrow f (- x) = - f (x)$

So, $f (x)$ is an odd function.

$∴ I = 0$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Properties of definite integration

⏱ 2mℹ️ Source: PYQ_2021

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