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

The integral of e^xln(e^x+1) is e^xln(e^x+1)+ln(e^x+1)-e^x-1+C

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integral of e^xln(e^x+1)

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

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

Solution steps

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

Apply u-substitution

: \int ln (u) d u

= \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= u ln (u) - \int 1 d u

\int 1 d u = u

= u ln (u) - u

Substitute back

u = e^{x} + 1

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

Simplify

(e^{x} + 1) ln (e^{x} + 1) - (e^{x} + 1) : e^{x} ln (e^{x} + 1) + ln (e^{x} + 1) - e^{x} - 1

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

Add a constant to the solution

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

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