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प्रश्न
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
बेरीज
उत्तर
\[\int\left( \frac{e^{m \tan^{- 1} x}}{1 + x^2} \right)dx\]
\[\text{Let} \tan^{- 1} x = t\]
\[ \Rightarrow \left( \frac{1}{1 + x^2} \right)dx = dt\]
\[Now, \int\left( \frac{e^{m \tan^{- 1} x}}{1 + x^2} \right)dx\]
\[ = \int e^{mt} dt\]
\[ = \frac{e^{mt}}{m} + C\]
\[ = \frac{e^{m \tan^{- 1} x}}{m} + C\]
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