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

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

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

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

ln ∣ ln (e^{x} + 1) ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ln ( u )} d u

= \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln ∣ ln (e^{x} + 1) ∣

Add a constant to the solution

= ln ∣ ln (e^{x} + 1) ∣ + C

y = ln (x^{2}), at x = 3

\int sinh^{2} (x) d x

\int_{0}^{4} x (x + 2) d x

x^{^{''}} + 5 x^{^{'}} = 7

\int \frac{4}{t + 1} d t

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