What is the integral of 1/((e^x+1)^2e^{-x)} ?

The integral of 1/((e^x+1)^2e^{-x)} is -1/(e^x+1)+C

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integral of 1/((e^x+1)^2e^{-x)}

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

- \frac{1}{e ^{x} + 1} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = e^{x} + 1

= - \frac{1}{e ^{x} + 1}

Add a constant to the solution

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

\int \frac{x ^{2}}{2} \frac{1}{1 + x ^{2}} d x

\int (2 x^{2} - x + 3) d x

\int \frac{x ^{2} - x + 5}{x + 3} d x

\int \frac{x - x ^{- 3}}{x ^{2} + x ^{- 2}} d x

\int (3 x^{2} + 4 x - 5) d x

What is the integral of 1/((e^x+1)^2e^{-x)} ?
The integral of 1/((e^x+1)^2e^{-x)} is -1/(e^x+1)+C