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प्रश्न
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
बेरीज
उत्तर
\[\int\frac{e^x dx}{\sqrt{16 - \left( e^x \right)^2}}\]
\[\text{ let } e^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[Now, \int\frac{e^x dx}{\sqrt{16 - \left( e^x \right)^2}}\]
\[ = \int\frac{dt}{\sqrt{16 - t^2}}\]
\[ = \int\frac{dt}{\sqrt{4^2 - t^2}}\]
\[ = \sin^{- 1} \left( \frac{t}{4} \right) + C\]
\[ = \sin^{- 1} \left( \frac{e^x}{4} \right) + C\]
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