What is the integral of (16)/(16+e^x) ?

The integral of (16)/(16+e^x) is x-ln|e^x+16|+C

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integral of (16)/(16+e^x)

\int \frac{16}{16 + e ^{x}} d x

x - ln ∣ e^{x} + 16 ∣ + C

Solution steps

\int \frac{16}{16 + e ^{x}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{16 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 16 + u )} d u

= 16 \cdot \int \frac{1}{u ( 16 + u )} d u

Take the partial fraction of

\frac{1}{u ( 16 + u )} : \frac{1}{16 u} - \frac{1}{16 ( u + 16 )}

= 16 \cdot \int \frac{1}{16 u} - \frac{1}{16 ( u + 16 )} 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

= 16 (\int \frac{1}{16 u} d u - \int \frac{1}{16 ( u + 16 )} d u)

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

\int \frac{1}{16 ( u + 16 )} d u = \frac{1}{16} ln ∣ u + 16 ∣

= 16 (\frac{1}{16} ln ∣ u ∣ - \frac{1}{16} ln ∣ u + 16 ∣)

Substitute back

u = e^{x}

= 16 (\frac{1}{16} ln ∣ e^{x} ∣ - \frac{1}{16} ln ∣ e^{x} + 16 ∣)

Simplify

16 (\frac{1}{16} ln ∣ e^{x} ∣ - \frac{1}{16} ln ∣ e^{x} + 16 ∣) : x - ln ∣ e^{x} + 16 ∣

= x - ln ∣ e^{x} + 16 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 16 ∣ + C

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

n = 1 \sum \infty \frac{( ( - 1 ) ^{n} )}{n}

\int \frac{v + 1}{v ^{2} + 1} d v

\int \frac{10}{( x ^{2} + 9 ) ( x - 1 )} d x

g (x) = tan (2 x) \cdot sin (2 x)

What is the integral of (16)/(16+e^x) ?
The integral of (16)/(16+e^x) is x-ln|e^x+16|+C