lim-x-0-x-arctan-t-2-dt-1-x-2-

Question Number 160218 by qaz last updated on 26/Nov/21

\lim_{x \to + \infty} \frac{\int_{0}^{x} {(\arctan t)}^{2} dt}{\sqrt{1 + x^{2}}} = ?

Answered by mathmax by abdo last updated on 26/Nov/21

= \lim_{x \to + \infty} \frac{\arctan^{2} x}{\frac{2 x}{2 \sqrt{1 + x^{2}}}} = \lim_{x \to + \infty} \arctan^{2} x \times \frac{\sqrt{1 + x^{2}}}{x}

= \lim_{x \to + \infty} \arctan^{2} x \times \sqrt{\frac{1 + x^{2}}{x^{2}}} = \lim_{x \to + \infty} \arctan^{2} x . \sqrt{1 + \frac{1}{x^{2}}}

= \frac{π^{2}}{4}