Question
Evalute : $\int \frac{x^3}{\sqrt{1+x^4}} d x$

Answer

Let $I=\int \frac{x^3}{\sqrt{1+x^4}} d x$
Put $1+x^4=t \quad \therefore 4 x^3 d x=d t$
$
\begin{aligned}
& \therefore x^3 d x=\frac{d t}{4} \\
& \begin{aligned}
\therefore I & =\int \frac{1}{\sqrt{t}} \cdot \frac{d t}{4}=\frac{1}{4} \int t^{-\frac{1}{2}} d t \\
& =\frac{1}{4} \cdot \frac{t^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c \\
& =\frac{1}{2} \sqrt{1+x^4}+c .
\end{aligned}
\end{aligned}
$

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