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Definite Integration question

2021 · 25 Jul · Shift 2 · Q30

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JEE Main
Mathematics
Definite Integration
2021 · 25 Jul · Shift 2 · Q30

Definite Integration question

2021 · 25 Jul · Shift 2 · Q30

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

The value of the integral

\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx}

is :

A
2
B
0
C
$-$ 1
D
1

View written solutionFree

Correct answer: B

Let $I=\int_{-1}^{1} \log\left(x+\sqrt{x^2+1}\right)\,dx.$
Observe the key identity: $\log\left(x+\sqrt{x^2+1}\right)=\sinh^{-1}x.$ So the integrand is $f(x)=\sinh^{-1}x.$
Check whether the function is odd: $f(-x)=\log\left(-x+\sqrt{x^2+1}\right).$ Now use $\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=1,$ so $\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}.$ Hence,

=\log\left(\frac{1}{\sqrt{x^2+1}+x}\right) =-\log\left(\sqrt{x^2+1}+x\right) =-f(x).$$ Therefore, $f(x)$ is an odd function. 4. Since $f(x)$ is odd and the limits are symmetric about $0$, we get $$I=\int_{-1}^{1} f(x)\,dx=0.$$ 5. Therefore, the value of the integral is $$\boxed{0}.$$ 6. Comparing with the stored correct answer: stored answer is **B**, which corresponds to $0$. This matches our result.

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