What is the integral of 9/(9+e^x) ?

The integral of 9/(9+e^x) is x-ln|e^x+9|+C

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integral of 9/(9+e^x)

\int \frac{9}{9 + e ^{x}} d x

x - ln ∣ e^{x} + 9 ∣ + C

Solution steps

\int \frac{9}{9 + e ^{x}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{9 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 9 + u )} d u

= 9 \cdot \int \frac{1}{u ( 9 + u )} d u

Take the partial fraction of

\frac{1}{u ( 9 + u )} : \frac{1}{9 u} - \frac{1}{9 ( u + 9 )}

= 9 \cdot \int \frac{1}{9 u} - \frac{1}{9 ( u + 9 )} 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

= 9 (\int \frac{1}{9 u} d u - \int \frac{1}{9 ( u + 9 )} d u)

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u + 9 )} d u = \frac{1}{9} ln ∣ u + 9 ∣

= 9 (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

Substitute back

u = e^{x}

= 9 (\frac{1}{9} ln ∣ e^{x} ∣ - \frac{1}{9} ln ∣ e^{x} + 9 ∣)

Simplify

9 (\frac{1}{9} ln ∣ e^{x} ∣ - \frac{1}{9} ln ∣ e^{x} + 9 ∣) : x - ln ∣ e^{x} + 9 ∣

= x - ln ∣ e^{x} + 9 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 9 ∣ + C

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