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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{{{({{\tan }^{ - 1}}x)}^3}}}{{1 + {x^2}}}\,dx = } $
  • A
    ${({\tan ^{ - 1}}x)^4} + c$
  • ✓
    $\frac{{{{({{\tan }^{ - 1}}x)}^4}}}{4} + c$
  • C
    $2{\tan ^{ - 1}}x + c$
  • D
    $2{({\tan ^{ - 1}}x)^2} + c$
✓

Answer

Correct option: B.
$\frac{{{{({{\tan }^{ - 1}}x)}^4}}}{4} + c$
b
(b) Put ${\tan ^{ - 1}}x = t \Rightarrow \frac{1}{{1 + {x^2}}}dx = dt$
 $\int_{}^{} {\frac{{{{({{\tan }^{ - 1}}x)}^3}}}{{1 + {x^2}}}} d\alpha = \int_{}^{} {{t^3}} dt = \frac{{{t^4}}}{4} + c$ = $\frac{{{{({{\tan }^{ - 1}}x)}^4}}}{4} + c$.

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