MCQ
$\int\limits_1^e {\left( {\frac{{{{\tan }^{ - 1}}x}}{x} + \frac{{\ln x}}{{\left( {1 + {x^2}} \right)}}} \right)} \,dx$ is equal to
  • A
    $\frac{1}{e}{\tan ^{ - 1}}e$
  • $tan^{-1}e$
  • C
    $e\ {\tan ^{ - 1}}\left( {\frac{1}{e}} \right)$
  • D
    $tan^{-1}(lne)$

Answer

Correct option: B.
$tan^{-1}e$
b
$\int_1^e {\mathop {{{\tan }^{ - 1}}}\limits_I } x \cdot \mathop {\frac{1}{x}}\limits_{II} dx + \int_1^e {\frac{{\ln x}}{{\left( {1 + {x^2}} \right)}}} dx$

$=\left(\tan ^{-1} x . \ln x\right)_{1}^{e}-\int_{1}^{e} \frac{1}{\left(1+x^{2}\right)} \ln x d x+\int_{1}^{e} \frac{\ln x}{\left(1+x^{2}\right)} d x$

$ = {\tan ^{ - 1}}(e) \cdot \ln e - {\tan ^{ - 1}}(1) \cdot \ln 1$

$=\tan ^{-1}(e)$

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