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