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∫ 1 ( X − 1 ) √ X + 2 D X - Mathematics

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प्रश्न

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

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

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अध्याय 19: Indefinite Integrals - Exercise 19.32 [पृष्ठ १९६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.32 | Q 1 | पृष्ठ १९६

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