What is the integral of (37)/(37+e^x) ?

The integral of (37)/(37+e^x) is x-ln|e^x+37|+C

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integral of (37)/(37+e^x)

\int \frac{37}{37 + e ^{x}} d x

x - ln ∣ e^{x} + 37 ∣ + C

Solution steps

\int \frac{37}{37 + e ^{x}} d x

Take the constant out:

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

= 37 \cdot \int \frac{1}{37 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 37 + u )} d u

= 37 \cdot \int \frac{1}{u ( 37 + u )} d u

Take the partial fraction of

\frac{1}{u ( 37 + u )} : \frac{1}{37 u} - \frac{1}{37 ( u + 37 )}

= 37 \cdot \int \frac{1}{37 u} - \frac{1}{37 ( u + 37 )} 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

= 37 (\int \frac{1}{37 u} d u - \int \frac{1}{37 ( u + 37 )} d u)

\int \frac{1}{37 u} d u = \frac{1}{37} ln ∣ u ∣

\int \frac{1}{37 ( u + 37 )} d u = \frac{1}{37} ln ∣ u + 37 ∣

= 37 (\frac{1}{37} ln ∣ u ∣ - \frac{1}{37} ln ∣ u + 37 ∣)

Substitute back

u = e^{x}

= 37 (\frac{1}{37} ln ∣ e^{x} ∣ - \frac{1}{37} ln ∣ e^{x} + 37 ∣)

Simplify

37 (\frac{1}{37} ln ∣ e^{x} ∣ - \frac{1}{37} ln ∣ e^{x} + 37 ∣) : x - ln ∣ e^{x} + 37 ∣

= x - ln ∣ e^{x} + 37 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 37 ∣ + C

x y^{^{'}} + y - y^{2} e^{2 x} = 0

f (x) = 7 ln (x)

\frac{d y}{d x} + y cos (x) = 7 cos (x), y (0) = 9

2 x^{2} - 5 x - 3

\int_{- 3}^{x} (t + 3) d t

What is the integral of (37)/(37+e^x) ?
The integral of (37)/(37+e^x) is x-ln|e^x+37|+C