Question

\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]

Answer 1

\[\text{ We  have, }\]
\[I = \int \frac{dx}{\left( x - 1 \right) \sqrt{2x + 3}}\]
\[\text{ Putting 2x} + 3 = t^2 \]
\[ \Rightarrow x = \frac{t^2 - 3}{2}\]
\[\text{ Diff both sides}\]
\[dx = t \text{ dt}\]
\[ \therefore I = \int \frac{t dt}{\left[ \frac{t^2 - 3}{2} - 1 \right]t}\]
\[ = \int\frac{2 \text{ dt }}{t^2 - 3 - 2}\]
\[ = \frac{2\text{ dt}}{t^2 - 5}\]
\[ = 2\int\frac{dt}{t^2 - \left( \sqrt{5} \right)^2}\]
\[ = 2 \times \frac{1}{2\sqrt{5}}\text{ log } \left| \frac{t - \sqrt{5}}{t + \sqrt{5}} \right| + C\]
\[ = \frac{1}{\sqrt{5}}\text{ log }\left| \frac{\sqrt{\text{ 2x + 3}} - \sqrt{5}}{\sqrt{2x + 3} + \sqrt{5}} \right| + C\]