What is the integral of (39)/(39+e^x) ?

The integral of (39)/(39+e^x) is x-ln|e^x+39|+C

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integral of (39)/(39+e^x)

\int \frac{39}{39 + e ^{x}} d x

x - ln ∣ e^{x} + 39 ∣ + C

Solution steps

\int \frac{39}{39 + e ^{x}} d x

Take the constant out:

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

= 39 \cdot \int \frac{1}{39 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 39 + u )} d u

= 39 \cdot \int \frac{1}{u ( 39 + u )} d u

Take the partial fraction of

\frac{1}{u ( 39 + u )} : \frac{1}{39 u} - \frac{1}{39 ( u + 39 )}

= 39 \cdot \int \frac{1}{39 u} - \frac{1}{39 ( u + 39 )} 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

= 39 (\int \frac{1}{39 u} d u - \int \frac{1}{39 ( u + 39 )} d u)

\int \frac{1}{39 u} d u = \frac{1}{39} ln ∣ u ∣

\int \frac{1}{39 ( u + 39 )} d u = \frac{1}{39} ln ∣ u + 39 ∣

= 39 (\frac{1}{39} ln ∣ u ∣ - \frac{1}{39} ln ∣ u + 39 ∣)

Substitute back

u = e^{x}

= 39 (\frac{1}{39} ln ∣ e^{x} ∣ - \frac{1}{39} ln ∣ e^{x} + 39 ∣)

Simplify

39 (\frac{1}{39} ln ∣ e^{x} ∣ - \frac{1}{39} ln ∣ e^{x} + 39 ∣) : x - ln ∣ e^{x} + 39 ∣

= x - ln ∣ e^{x} + 39 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 39 ∣ + C

x \to - 2 lim (5 x + 2)

y = x

y^{^{''}} - 2 y^{^{'}} + y = 14.5 e^{t}

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

y^{^{'}} + \frac{y}{x} = 9 x^{3} y^{2}

What is the integral of (39)/(39+e^x) ?
The integral of (39)/(39+e^x) is x-ln|e^x+39|+C