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

The integral of ((e^xln(e^x+1)))/(e^x+1) is (ln^2(e^x+1))/2+C

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

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

\frac{ln ^{2} ( e ^{x} + 1 )}{2} + C

Solution steps

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

Apply u-substitution

= \int \frac{ln ( u )}{u} d u

Apply u-substitution

= \int v d v

Apply the Power Rule

= \frac{v ^{2}}{2}

Substitute back

= \frac{ln ^{2} ( e ^{x} + 1 )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( e ^{x} + 1 )}{2} + C

What is the integral of ((e^xln(e^x+1)))/(e^x+1) ?
The integral of ((e^xln(e^x+1)))/(e^x+1) is (ln^2(e^x+1))/2+C