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Question
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Solution
`\text{ We have,} `
\[I = \int \frac{x dx}{\left( x - 3 \right) \sqrt{x + 1}}\]
\[\text{ Putting x }+ 1 = t^2 \]
\[ \Rightarrow x = t^2 - 1\]
\[\text{ Diff both sides }\]
\[dx = 2t \text{ dt }\]
\[ \therefore I = \int \frac{\left( t^2 - 1 \right)2t \text{ dt }}{\left( t^2 - 1 - 3 \right)t}\]
\[ = 2\int \left( \frac{t^2 - 1}{t^2 - 4} \right)dt\]
\[ = 2\int\left( \frac{t^2 - 4 + 3}{t^2 - 4} \right)dt\]
\[ = 2\int\left( \frac{t^2 - 4}{t^2 - 4} \right)dt + 6\int \frac{dt}{t^2 - 2^2}\]
\[ = 2\int dt + 6\int\frac{dt}{t^2 - 2^2}\]
\[ = 2t + 6 \times \frac{1}{2 \times 2}\text{ log }\left| \frac{t - 2}{t + 2} \right| + C\]
\[ = 2\sqrt{x + 1} + \frac{3}{2}\text{ log} \left| \frac{t - 2}{t + 2} \right| + C\]
\[ = 2\sqrt{x + 1} + \frac{3}{2}\text{ log }\left| \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} \right| + C\]
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