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प्रश्न

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

\[\int\frac{dx}{\sqrt{\left( 2 - x \right)^2 - 1}}\]
\[\text{ let 2 }- x = t\]
\[ \Rightarrow - dx = dt\]
\[ \Rightarrow dx = - dt\]
\[Now, \int\frac{dx}{\sqrt{\left( 2 - x \right)^2 - 1}}\]
\[ = \int\frac{- dt}{\sqrt{t^2 - 1}}\]
\[ = - \text{ log }\left| t + \sqrt{t^2 - 1} \right| + C\]
\[ = - \text{ log }\left| \left( 2 - x \right) + \sqrt{\left( 2 - x \right)^2 - 1} \right| + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.14 [पृष्ठ ८३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.14 | Q 9 | पृष्ठ ८३

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