Question
Evalute : $\int \frac{x^5}{x^2+1} d x$

Answer

Let $I=\int \frac{x^5}{x^2+1} d x=\int \frac{x^4}{x^2+1} \cdot x d x$
$
=\int \frac{\left(x^2\right)^2}{x^2+1} \cdot x d x
$
Put $x^2+1=t \quad \therefore 2 x d x=d t$
$
\begin{aligned}
& \therefore x d x=\frac{d t}{2} \text { and } x^2=t-1 \\
& \begin{aligned}
\therefore I & =\int \frac{(t-1)^2}{t} \cdot \frac{d t}{2}=\frac{1}{2} \int\left(\frac{t^2-2 t+1}{t}\right) d t \\
& =\frac{1}{2} \int\left(t-2+\frac{1}{t}\right) d t \\
& =\frac{1}{2} \int t d t-\int 1 d t+\frac{1}{2} \int \frac{1}{t} d t \\
& =\frac{1}{2} \cdot \frac{t^2}{2}-t+\frac{1}{2} \log |t|+c
\end{aligned}
\end{aligned}
$
$
=\frac{1}{4}\left(x^2+1\right)^2-\left(x^2+1\right)+\frac{1}{2} \log \left|x^2+1\right|+c .$

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