What is the integral of (38)/(38+e^x) ?

The integral of (38)/(38+e^x) is x-ln|e^x+38|+C

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integral of (38)/(38+e^x)

\int \frac{38}{38 + e ^{x}} d x

x - ln ∣ e^{x} + 38 ∣ + C

Solution steps

\int \frac{38}{38 + e ^{x}} d x

Take the constant out:

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

= 38 \cdot \int \frac{1}{38 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 38 + u )} d u

= 38 \cdot \int \frac{1}{u ( 38 + u )} d u

Take the partial fraction of

\frac{1}{u ( 38 + u )} : \frac{1}{38 u} - \frac{1}{38 ( u + 38 )}

= 38 \cdot \int \frac{1}{38 u} - \frac{1}{38 ( u + 38 )} 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

= 38 (\int \frac{1}{38 u} d u - \int \frac{1}{38 ( u + 38 )} d u)

\int \frac{1}{38 u} d u = \frac{1}{38} ln ∣ u ∣

\int \frac{1}{38 ( u + 38 )} d u = \frac{1}{38} ln ∣ u + 38 ∣

= 38 (\frac{1}{38} ln ∣ u ∣ - \frac{1}{38} ln ∣ u + 38 ∣)

Substitute back

u = e^{x}

= 38 (\frac{1}{38} ln ∣ e^{x} ∣ - \frac{1}{38} ln ∣ e^{x} + 38 ∣)

Simplify

38 (\frac{1}{38} ln ∣ e^{x} ∣ - \frac{1}{38} ln ∣ e^{x} + 38 ∣) : x - ln ∣ e^{x} + 38 ∣

= x - ln ∣ e^{x} + 38 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 38 ∣ + C

\int \frac{3 + u}{1 - u ^{2}} d u

x \to - 5 lim (x^{2} + 5)

\frac{d y}{d x} = e^{y - 2 x}

What is the integral of (38)/(38+e^x) ?
The integral of (38)/(38+e^x) is x-ln|e^x+38|+C