What is the integral of (24)/(24+e^x) ?

The integral of (24)/(24+e^x) is x-ln|e^x+24|+C

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integral of (24)/(24+e^x)

\int \frac{24}{24 + e ^{x}} d x

x - ln ∣ e^{x} + 24 ∣ + C

Solution steps

\int \frac{24}{24 + e ^{x}} d x

Take the constant out:

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

= 24 \cdot \int \frac{1}{24 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 24 + u )} d u

= 24 \cdot \int \frac{1}{u ( 24 + u )} d u

Take the partial fraction of

\frac{1}{u ( 24 + u )} : \frac{1}{24 u} - \frac{1}{24 ( u + 24 )}

= 24 \cdot \int \frac{1}{24 u} - \frac{1}{24 ( u + 24 )} 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

= 24 (\int \frac{1}{24 u} d u - \int \frac{1}{24 ( u + 24 )} d u)

\int \frac{1}{24 u} d u = \frac{1}{24} ln ∣ u ∣

\int \frac{1}{24 ( u + 24 )} d u = \frac{1}{24} ln ∣ u + 24 ∣

= 24 (\frac{1}{24} ln ∣ u ∣ - \frac{1}{24} ln ∣ u + 24 ∣)

Substitute back

u = e^{x}

= 24 (\frac{1}{24} ln ∣ e^{x} ∣ - \frac{1}{24} ln ∣ e^{x} + 24 ∣)

Simplify

24 (\frac{1}{24} ln ∣ e^{x} ∣ - \frac{1}{24} ln ∣ e^{x} + 24 ∣) : x - ln ∣ e^{x} + 24 ∣

= x - ln ∣ e^{x} + 24 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 24 ∣ + C

x \to 4 - lim ([x] - x)

\int (x^{2} + 1)^{4} d x

{\frac{1}{s ^{3} + 5 s}}

\int e^{- 3 x} d x

\int \frac{41}{x ^{3} - 27} d x

What is the integral of (24)/(24+e^x) ?
The integral of (24)/(24+e^x) is x-ln|e^x+24|+C