The value of limx→∞ ∫0x tan−1⁡x2dxx2+1 is

limx→∞ ∫0x tan−1⁡x2dxx2+1=limx→∞ tan−1⁡x2x2+1x =limx→∞ tan−1⁡x21+1x2=π22=π24.

The value of limx→∞ ∫0x tan−1⁡x2dxx2+1 is

limx→∞ ∫0x tan−1⁡x2dxx2+1=limx→∞ tan−1⁡x2x2+1x =limx→∞ tan−1⁡x21+1x2=π22=π24.

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The value of $lim_{x \to \infty} \frac{\int_{0}^{x} {(\tan^{- 1} x)}^{2} d x}{\sqrt{x^{2} + 1}}$ is

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$π / 4$

$π^{2} / 2$

$π^{2} / 4$

$π$

answer is C.

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$lim_{x \to \infty} \frac{\int_{0}^{x} {(\tan^{- 1} x)}^{2} d x}{\sqrt{x^{2} + 1}} = lim_{x \to \infty} \frac{{(\tan^{- 1} x)}^{2} \sqrt{x^{2} + 1}}{x}$

$= lim_{x \to \infty} {(\tan^{- 1} x)}^{2} \sqrt{1 + \frac{1}{x^{2}}} = {(\frac{π}{2})}^{2} = \frac{π^{2}}{4} .$

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The value of limx→∞ ∫0x tan−1⁡x2dxx2+1 is

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The value of $lim_{x \to \infty} \frac{\int_{0}^{x} {(\tan^{- 1} x)}^{2} d x}{\sqrt{x^{2} + 1}}$ is