I would like to know how to find u as shown below.
This question is an indefinite integral, and it is a whole question. No method specified.
, where a is a constant.
Hint
$\displaystyle \int {x\sqrt {\frac{{1 - a{x^2}}}{{1 - {x^2}}}} dx} = \left\{ \begin{gathered}
\frac{{1 - a{x^2}}}
{{1 - {x^2}}} = {t^2} \Leftrightarrow {x^2} = \frac{{{t^2} - 1}}
{{{t^2} - a}}, \hfill \\
xdx = \frac{{\left( {1 - a} \right)t}}
{{{{\left( {{t^2} - a} \right)}^2}}}dt \hfill \\
\end{gathered} \right\} = \left( {1 - a} \right)\int {\frac{{{t^2}}}
{{{{\left( {{t^2} - a} \right)}^2}}}dt}$
$\displaystyle = \frac{{a - 1}}
{2}\int {td\left( {\frac{1}
{{{t^2} - a}}} \right)} = \frac{{\left( {a - 1} \right)t}}
{{2\left( {{t^2} - a} \right)}} - \frac{{a - 1}}
{2}\int {\frac{{dt}}
{{{t^2} - a}}}$
Now consider three cases:
1) $\displaystyle a<0$;
2) $\displaystyle a=0$;
3) $\displaystyle a>0$.