(D) We know that for $y > 0$,$\tan ^{-1}(y) + \tan ^{-1}(1/y) = \frac{\pi}{2}$.Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.Therefore,the i

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(D) We know that for $y > 0$,$\tan ^{-1}(y) + \tan ^{-1}(1/y) = \frac{\pi}{2}$.Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.Therefore,the integrand simplifies as follows:$\tan ^{-1}\left(\frac{x}{x^2+1}\right) + \tan ^{-1}\left(\frac{x^2+1}{x}\right) = \frac{\pi}{2}$.Now,evaluate the integral:$I = \int_1^2 \frac{\pi}{2} d x$$I = \frac{\pi}{2} [x]_1^2$$I = \frac{\pi}{2} (2 - 1) = \frac{\pi}{2}$.

Class 12 Mathematics 7-2.Definite Integral Fundamental definite integration

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$\int_1^2 \left( \tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right) \right) d x =$

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A
$\frac{\pi}{4}$
B
$\frac{3 \pi}{4}$
C
$\frac{\pi}{4}$
D
$\frac{\pi}{2}$

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7-2.Definite Integral

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If $I$ is the greatest of $I_1=\int_0^1 e^{-x} \cos ^2 x \, dx, I_2=\int_0^1 e^{-x^2} \cos ^2 x \, dx, I_3=\int_0^1 e^{-x^2} \, dx, I_4=\int_0^1 e^{-x^2 / 2} \, dx$, then

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$\int_1^2 \left( \tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right) \right) d x =$

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