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

Answer

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

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