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The value of $\int \frac{e^x}{e^x+1} d x$ is
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Verified Answer
The correct answer is:
$\log \left(e^x+1\right)+c$
\(\begin{aligned}
& \Rightarrow \int \frac{e^x}{e^x+1} \cdot d x \\
& \text {Let, } e^x+1=t \\
& \text {So, } d x \cdot e^x=d t \\
& \Rightarrow \int \frac{d t}{t} \Rightarrow \log (t)+C \\
& \Rightarrow \log \left(e^x+1\right)+c
\end{aligned}\)
& \Rightarrow \int \frac{e^x}{e^x+1} \cdot d x \\
& \text {Let, } e^x+1=t \\
& \text {So, } d x \cdot e^x=d t \\
& \Rightarrow \int \frac{d t}{t} \Rightarrow \log (t)+C \\
& \Rightarrow \log \left(e^x+1\right)+c
\end{aligned}\)
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