What is the integral of 5/(5+e^x) ?

The integral of 5/(5+e^x) is x-ln|e^x+5|+C

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integral of 5/(5+e^x)

\int \frac{5}{5 + e ^{x}} d x

x - ln ∣ e^{x} + 5 ∣ + C

Solution steps

\int \frac{5}{5 + e ^{x}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{5 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 5 + u )} d u

= 5 \cdot \int \frac{1}{u ( 5 + u )} d u

Take the partial fraction of

\frac{1}{u ( 5 + u )} : \frac{1}{5 u} - \frac{1}{5 ( u + 5 )}

= 5 \cdot \int \frac{1}{5 u} - \frac{1}{5 ( u + 5 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int \frac{1}{5 u} d u - \int \frac{1}{5 ( u + 5 )} d u)

\int \frac{1}{5 u} d u = \frac{1}{5} ln ∣ u ∣

\int \frac{1}{5 ( u + 5 )} d u = \frac{1}{5} ln ∣ u + 5 ∣

= 5 (\frac{1}{5} ln ∣ u ∣ - \frac{1}{5} ln ∣ u + 5 ∣)

Substitute back

u = e^{x}

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

Simplify

5 (\frac{1}{5} ln ∣ e^{x} ∣ - \frac{1}{5} ln ∣ e^{x} + 5 ∣) : x - ln ∣ e^{x} + 5 ∣

= x - ln ∣ e^{x} + 5 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 5 ∣ + C

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