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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
$\int_{}^{} {\frac{{3{x^2}}}{{{x^6} + 1}}dx = } $
  • A
    $\log ({x^6} + 1) + c$
  • ✓
    ${\tan ^{ - 1}}({x^3}) + c$
  • C
    $3{\tan ^{ - 1}}({x^3}) + c$
  • D
    $3{\tan ^{ - 1}}\left( {\frac{{{x^3}}}{3}} \right) + c$
✓

Answer

Correct option: B.
${\tan ^{ - 1}}({x^3}) + c$
b
(b) Put ${x^3} = t \Rightarrow 3{x^2}dx = dt,$ therefore
$\int_{}^{} {\frac{{3{x^2}}}{{{x^6} + 1}}\,dx = \int_{}^{} {\frac{1}{{{t^2} + 1}}dt = {{\tan }^{ - 1}}(t) + c} }  = {\tan ^{ - 1}}({x^3}) + c$.

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