What is the integral of (e^{3-x})/((1+e^{3-x))^2} ?

The integral of (e^{3-x})/((1+e^{3-x))^2} is 1/(1+e^{3-x)}+C

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integral of (e^{3-x})/((1+e^{3-x))^2}

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

\frac{1}{1 + e ^{3 - x}} + C

Solution steps

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

Apply u-substitution

= \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})

Substitute back

u = 1 + e^{3 - x}

= - (- \frac{1}{1 + e ^{3 - x}})

Simplify

= \frac{1}{1 + e ^{3 - x}}

Add a constant to the solution

= \frac{1}{1 + e ^{3 - x}} + C

What is the integral of (e^{3-x})/((1+e^{3-x))^2} ?
The integral of (e^{3-x})/((1+e^{3-x))^2} is 1/(1+e^{3-x)}+C