What is the integral of (e^x)/((e^x+1)^6) ?

The integral of (e^x)/((e^x+1)^6) is -1/(5(e^x+1)^5)+C

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integral of (e^x)/((e^x+1)^6)

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{6}} d x

- \frac{1}{5 ( e ^{x} + 1 ) ^{5}} + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{6}} d x

Apply u-substitution

: \int \frac{1}{u ^{6}} d u

= \int \frac{1}{u ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= - \frac{1}{5 u ^{5}}

Substitute back

u = e^{x} + 1

= - \frac{1}{5 ( e ^{x} + 1 ) ^{5}}

Add a constant to the solution

= - \frac{1}{5 ( e ^{x} + 1 ) ^{5}} + C

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What is the integral of (e^x)/((e^x+1)^6) ?
The integral of (e^x)/((e^x+1)^6) is -1/(5(e^x+1)^5)+C