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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: - \frac{1}{u}

= - (- \frac{1}{u})

Substitute back

u = e^{- x} + x

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

Simplify

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

Add a constant to the solution

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

\int sin (x) (sec^{2} (x) + cot^{2} (x)) d x

\int \frac{x ^{2} + 2}{x ^{2} - 4} d x

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

\int \frac{x + 1}{x ^{2} - 5 x + 6} d x

\int \frac{x ^{3} arccos ( x )}{1 - x ^{2}} d x

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