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