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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
The value of $\int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 + {x^4}} }}\;dx} $ is
  • A
    ${(1 + {x^4})^{\frac{1}{2}}} + c$
  • B
    $ - {(1 + {x^4})^{\frac{1}{2}}} + c$
  • ✓
    $\frac{1}{2}{(1 + {x^4})^{\frac{1}{2}}} + c$
  • D
    $ - \frac{1}{2}{(1 + {x^4})^{\frac{1}{2}}} + c$
✓

Answer

Correct option: C.
$\frac{1}{2}{(1 + {x^4})^{\frac{1}{2}}} + c$
c
(c)$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 + {x^4}} }}\,dx} = \frac{1}{4}\int_{}^{} {\frac{{4{x^3}}}{{\sqrt {1 + {x^4}} }}\,dx} $

$({\rm{Put}}\,1 + {x^4} = t)$

$⇒$$4{x^3}dx = dt$

$ = \frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^{1/2}}}}} = \frac{1}{4}\frac{{{t^{ - \frac{1}{2} + 1}}}}{{ - \frac{1}{2} + 1}} = \frac{1}{2}\sqrt t = \frac{1}{2}\sqrt {1 + {x^4}} + c.$

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