MCQ
$\int_{}^{} {\frac{x}{{1 + {x^4}}}\;dx = } $
  • A
    $\frac{1}{2}{\cot ^{ - 1}}{x^2} + c$
  • $\frac{1}{2}{\tan ^{ - 1}}{x^2} + c$
  • C
    ${\cot ^{ - 1}}{x^2} + c$
  • D
    ${\tan ^{ - 1}}{x^2} + c$

Answer

Correct option: B.
$\frac{1}{2}{\tan ^{ - 1}}{x^2} + c$
b
(b) Put $t = {x^2} \Rightarrow dt = 2x\,dx,$ therefore
$\int_{}^{} {\frac{x}{{1 + {x^4}}}\,dx = \frac{1}{2}\int_{}^{} {\frac{1}{{1 + {t^2}}}\,dt = \frac{1}{2}{{\tan }^{ - 1}}t + c = \frac{1}{2}{{\tan }^{ - 1}}{x^2} + c} } $.

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