MCQ
$\int\limits_1^e {\left( {\frac{{{{\tan }^{ - 1}}x}}{x} + \frac{{\ln x}}{{\left( {1 + {x^2}} \right)}}} \right)} \,dx$ મેળવો.
- A$\frac{1}{e}{\tan ^{ - 1}}e$
- B$tan^{-1}e$
- C$e\ {\tan ^{ - 1}}\left( {\frac{1}{e}} \right)$
- D$tan^{-1}(lne)$
$=\left(\tan ^{-1} x . \ln x\right)_{1}^{e}-\int_{1}^{e} \frac{1}{\left(1+x^{2}\right)} \ln x d x+\int_{1}^{e} \frac{\ln x}{\left(1+x^{2}\right)} d x$
$ = {\tan ^{ - 1}}(e) \cdot \ln e - {\tan ^{ - 1}}(1) \cdot \ln 1$
$=\tan ^{-1}(e)$
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