Question
Evaluate the following.
$
\int \frac{x^3}{\sqrt{1+x^4}} d x
$

Answer

$
\text { Let } I =\int \frac{ x ^3}{\sqrt{1+ x ^4}} dx
$

Put $1+x^4=t$
$
\therefore 4 x ^3 \cdot dx = dt
$
$\therefore x ^3 \cdot dx =\frac{1}{4} dt$
$
\therefore I =\frac{1}{4} \int \frac{ dt }{\sqrt{ t }}
$
$=\frac{1}{4} \int t ^{\frac{-1}{2}} dt$
$
=\frac{1}{4} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+ c
$
$
=\frac{1}{2} \sqrt{t}+c
$
$\therefore I =\frac{1}{2} \sqrt{1+ x ^4}+ c$

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