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प्रश्न
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
बेरीज
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उत्तर
\[\text{Let I} = \int\frac{e^{3x}}{e^{3x} + 1}dx\]
\[\text{Putting }e^{3x} + 1 = t \]
\[ \Rightarrow 3 e^{3x} = \frac{dt}{dx}\]
\[ \Rightarrow dx = \frac{dt}{3 e^{3x}}\]
\[ \therefore I = \int\frac{e^{3x}}{3t\left( e^{3x} \right)}dt\]
\[ = \frac{1}{3}\int\frac{1}{t}dt\]
\[ = \frac{\text{ln }\left| t \right|}{3} + C\]
\[ = \frac{\text{ln} \left| e^{3x} + 1 \right|}{3} + C\]
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