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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

(2 x - 1) d x + (7 y + 9) d y = 0

(x^{3 x})^{^{'}}

(1 + cos (x)) y^{^{'}} = (1 + e^{- y}) sin (x)

\int \frac{1}{( 2 x - 1 ) ^{2}} d x

\frac{\partial}{\partial x} (arctan (x^{2} y))

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