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