First slide

Evaluation of definite integrals

Question

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

Moderate

A

$π / 4$
B

$π^{2} / 2$
C

$π^{2} / 4$
D

$π$

Solution

$lim_{x \to \infty} \frac{\int_{0}^{x} {(\tan^{- 1} x)}^{2} d x}{\sqrt{x^{2} + 1}} = lim_{x \to \infty} \frac{{(\tan^{- 1} x)}^{2} \sqrt{x^{2} + 1}}{x}$
(L’ Hôpital Rule)
$= lim_{x \to \infty} {(\tan^{- 1} x)}^{2} \sqrt{1 + \frac{1}{x^{2}}} = {(\frac{π}{2})}^{2} = \frac{π^{2}}{4}$

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