What is the integral of (ln(e^x+1))/(e^x) ?

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

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integral of (ln(e^x+1))/(e^x)

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply Integration By Parts

= - 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

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