MCQ
$\int_{\, - \,1}^{\,3} {\,{{\tan }^{ - 1}}\left( {\frac{x}{{{x^2} + 1}}} \right) + {{\tan }^{ - 1}}\left( {\frac{{{x^2} + 1}}{x}} \right)\,dx} =$
- ✓$2\pi $
- B$\pi $
- C$\frac{{21}}{5}\pi $
- D$\frac{\pi }{4}$
$ = \int_{ - 1}^3 {\left( {\frac{\pi }{2}} \right)\,dx= \left[ {\frac{{\pi \,x}}{2}} \right]_{ - 1}^3 = 2\pi } $,
$\left( \because {{\tan }^{-1}}(x)+{{\cot }^{-1}}(x)=\frac{\pi }{2} \right)$ .
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
{જ્યા $x,y \in R^+, x^2y + x \ne 0$ }
$\,\left| {\begin{array}{*{20}{c}} {\vec a .\,\vec a }&{\vec a .\vec b }&{\vec a .\vec c } \\ {\vec b .\vec a }&{\vec b .\vec b }&{\vec b .\vec c } \\ {\vec c .\vec a }&{\vec c .\vec b }&{\vec c .\vec c } \end{array}} \right|\,\, = \,\,..........$