What is the integral of (e^x)/((e^x+1)^{16)} ?

The integral of (e^x)/((e^x+1)^{16)} is -1/(15(e^x+1)^{15)}+C

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integral of (e^x)/((e^x+1)^{16)}

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{16}} d x

- \frac{1}{15 ( e ^{x} + 1 ) ^{15}} + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{16}} d x

Apply u-substitution

: \int \frac{1}{u ^{16}} d u

= \int \frac{1}{u ^{16}} d u

Apply the Power Rule

: - \frac{1}{15 u ^{15}}

= - \frac{1}{15 u ^{15}}

Substitute back

u = e^{x} + 1

= - \frac{1}{15 ( e ^{x} + 1 ) ^{15}}

Add a constant to the solution

= - \frac{1}{15 ( e ^{x} + 1 ) ^{15}} + C

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What is the integral of (e^x)/((e^x+1)^{16)} ?
The integral of (e^x)/((e^x+1)^{16)} is -1/(15(e^x+1)^{15)}+C