What is the integral of (29)/(29+e^x) ?

The integral of (29)/(29+e^x) is x-ln|e^x+29|+C

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integral of (29)/(29+e^x)

\int \frac{29}{29 + e ^{x}} d x

x - ln ∣ e^{x} + 29 ∣ + C

Solution steps

\int \frac{29}{29 + e ^{x}} d x

Take the constant out:

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

= 29 \cdot \int \frac{1}{29 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 29 + u )} d u

= 29 \cdot \int \frac{1}{u ( 29 + u )} d u

Take the partial fraction of

\frac{1}{u ( 29 + u )} : \frac{1}{29 u} - \frac{1}{29 ( u + 29 )}

= 29 \cdot \int \frac{1}{29 u} - \frac{1}{29 ( u + 29 )} 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

= 29 (\int \frac{1}{29 u} d u - \int \frac{1}{29 ( u + 29 )} d u)

\int \frac{1}{29 u} d u = \frac{1}{29} ln ∣ u ∣

\int \frac{1}{29 ( u + 29 )} d u = \frac{1}{29} ln ∣ u + 29 ∣

= 29 (\frac{1}{29} ln ∣ u ∣ - \frac{1}{29} ln ∣ u + 29 ∣)

Substitute back

u = e^{x}

= 29 (\frac{1}{29} ln ∣ e^{x} ∣ - \frac{1}{29} ln ∣ e^{x} + 29 ∣)

Simplify

29 (\frac{1}{29} ln ∣ e^{x} ∣ - \frac{1}{29} ln ∣ e^{x} + 29 ∣) : x - ln ∣ e^{x} + 29 ∣

= x - ln ∣ e^{x} + 29 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 29 ∣ + C

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What is the integral of (29)/(29+e^x) ?
The integral of (29)/(29+e^x) is x-ln|e^x+29|+C