What is the integral of (28)/(28+e^x) ?

The integral of (28)/(28+e^x) is x-ln|e^x+28|+C

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integral of (28)/(28+e^x)

\int \frac{28}{28 + e ^{x}} d x

x - ln ∣ e^{x} + 28 ∣ + C

Solution steps

\int \frac{28}{28 + e ^{x}} d x

Take the constant out:

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

= 28 \cdot \int \frac{1}{28 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 28 + u )} d u

= 28 \cdot \int \frac{1}{u ( 28 + u )} d u

Take the partial fraction of

\frac{1}{u ( 28 + u )} : \frac{1}{28 u} - \frac{1}{28 ( u + 28 )}

= 28 \cdot \int \frac{1}{28 u} - \frac{1}{28 ( u + 28 )} 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

= 28 (\int \frac{1}{28 u} d u - \int \frac{1}{28 ( u + 28 )} d u)

\int \frac{1}{28 u} d u = \frac{1}{28} ln ∣ u ∣

\int \frac{1}{28 ( u + 28 )} d u = \frac{1}{28} ln ∣ u + 28 ∣

= 28 (\frac{1}{28} ln ∣ u ∣ - \frac{1}{28} ln ∣ u + 28 ∣)

Substitute back

u = e^{x}

= 28 (\frac{1}{28} ln ∣ e^{x} ∣ - \frac{1}{28} ln ∣ e^{x} + 28 ∣)

Simplify

28 (\frac{1}{28} ln ∣ e^{x} ∣ - \frac{1}{28} ln ∣ e^{x} + 28 ∣) : x - ln ∣ e^{x} + 28 ∣

= x - ln ∣ e^{x} + 28 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 28 ∣ + C

2 x \frac{d y}{d x} + y = 6 x

x \to 3 + lim (6 + x)

y = \frac{6}{x ^{2}}, (1, 6)

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

y = 3 x^{3} - x^{2} + 5, (2, 25)

What is the integral of (28)/(28+e^x) ?
The integral of (28)/(28+e^x) is x-ln|e^x+28|+C