What is the integral of e^{-x}ln(e^x+1) ?

The integral of e^{-x}ln(e^x+1) is -e^{-x}ln(e^x+1)-ln|e^x+1|+x+C

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integral of e^{-x}ln(e^x+1)

\int e^{- x} ln (e^{x} + 1) d x

- e^{- x} ln (e^{x} + 1) - ln ∣ e^{x} + 1 ∣ + x + C

Solution steps

\int e^{- x} ln (e^{x} + 1) d x

Apply Integration By Parts

: - e^{- x} ln (e^{x} + 1) - \int - \frac{1}{e ^{x} + 1} d x

= - e^{- x} ln (e^{x} + 1) - \int - \frac{1}{e ^{x} + 1} d x

\int - \frac{1}{e ^{x} + 1} d x = ln ∣ e^{x} + 1 ∣ - x

= - e^{- x} ln (e^{x} + 1) - (ln ∣ e^{x} + 1 ∣ - x)

Simplify

= - e^{- x} ln (e^{x} + 1) - ln ∣ e^{x} + 1 ∣ + x

Add a constant to the solution

= - e^{- x} ln (e^{x} + 1) - ln ∣ e^{x} + 1 ∣ + x + C

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What is the integral of e^{-x}ln(e^x+1) ?
The integral of e^{-x}ln(e^x+1) is -e^{-x}ln(e^x+1)-ln|e^x+1|+x+C