What is the integral of (12)/(12+e^x) ?

The integral of (12)/(12+e^x) is x-ln|e^x+12|+C

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integral of (12)/(12+e^x)

\int \frac{12}{12 + e ^{x}} d x

x - ln ∣ e^{x} + 12 ∣ + C

Solution steps

\int \frac{12}{12 + e ^{x}} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{12 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 12 + u )} d u

= 12 \cdot \int \frac{1}{u ( 12 + u )} d u

Take the partial fraction of

\frac{1}{u ( 12 + u )} : \frac{1}{12 u} - \frac{1}{12 ( u + 12 )}

= 12 \cdot \int \frac{1}{12 u} - \frac{1}{12 ( u + 12 )} 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

= 12 (\int \frac{1}{12 u} d u - \int \frac{1}{12 ( u + 12 )} d u)

\int \frac{1}{12 u} d u = \frac{1}{12} ln ∣ u ∣

\int \frac{1}{12 ( u + 12 )} d u = \frac{1}{12} ln ∣ u + 12 ∣

= 12 (\frac{1}{12} ln ∣ u ∣ - \frac{1}{12} ln ∣ u + 12 ∣)

Substitute back

u = e^{x}

= 12 (\frac{1}{12} ln ∣ e^{x} ∣ - \frac{1}{12} ln ∣ e^{x} + 12 ∣)

Simplify

12 (\frac{1}{12} ln ∣ e^{x} ∣ - \frac{1}{12} ln ∣ e^{x} + 12 ∣) : x - ln ∣ e^{x} + 12 ∣

= x - ln ∣ e^{x} + 12 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 12 ∣ + C

\frac{t}{t + 1}

\int cot^{3} (11 x) d x

y = x^{3} - 3 x + 2 (2.4)

\int tan^{5} (π x) sec^{4} (π x) d x

x^{4} + 7 x^{2} - x

What is the integral of (12)/(12+e^x) ?
The integral of (12)/(12+e^x) is x-ln|e^x+12|+C