MCQ
$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 + {x^4}} }}\;dx} $ =
- 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$
$({\rm{Put}}\,1 + {x^4} = t)$
$⇒$$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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