MCQ
$\mathop {\lim }\limits_{x \to \infty } \frac{{\int\limits_0^x {{{\left( {{{\tan }^{ - 1}}x} \right)}^2}dx} }}{{\sqrt {{x^2} + 1} }} =\ ......$
- A$\frac{\pi }{4}$
- B$\pi $
- ✓$\frac{{{\pi ^2}}}{4}$
- D$\frac{{{\pi ^2}}}{2}$
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