What is the integral of (25)/(25+e^x) ?

The integral of (25)/(25+e^x) is x-ln|e^x+25|+C

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integral of (25)/(25+e^x)

\int \frac{25}{25 + e ^{x}} d x

x - ln ∣ e^{x} + 25 ∣ + C

Solution steps

\int \frac{25}{25 + e ^{x}} d x

Take the constant out:

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

= 25 \cdot \int \frac{1}{25 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 25 + u )} d u

= 25 \cdot \int \frac{1}{u ( 25 + u )} d u

Take the partial fraction of

\frac{1}{u ( 25 + u )} : \frac{1}{25 u} - \frac{1}{25 ( u + 25 )}

= 25 \cdot \int \frac{1}{25 u} - \frac{1}{25 ( u + 25 )} 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

= 25 (\int \frac{1}{25 u} d u - \int \frac{1}{25 ( u + 25 )} d u)

\int \frac{1}{25 u} d u = \frac{1}{25} ln ∣ u ∣

\int \frac{1}{25 ( u + 25 )} d u = \frac{1}{25} ln ∣ u + 25 ∣

= 25 (\frac{1}{25} ln ∣ u ∣ - \frac{1}{25} ln ∣ u + 25 ∣)

Substitute back

u = e^{x}

= 25 (\frac{1}{25} ln ∣ e^{x} ∣ - \frac{1}{25} ln ∣ e^{x} + 25 ∣)

Simplify

25 (\frac{1}{25} ln ∣ e^{x} ∣ - \frac{1}{25} ln ∣ e^{x} + 25 ∣) : x - ln ∣ e^{x} + 25 ∣

= x - ln ∣ e^{x} + 25 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 25 ∣ + C

x y^{^{'}} + y = - 3 x y^{2}, y (1) = - 4

f (x) = ln^{4} (x)

f (x) = \frac{1}{x - 8}

y = 4 - x^{2}

What is the integral of (25)/(25+e^x) ?
The integral of (25)/(25+e^x) is x-ln|e^x+25|+C