What is the integral of (40)/(40+e^x) ?

The integral of (40)/(40+e^x) is x-ln|e^x+40|+C

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integral of (40)/(40+e^x)

\int \frac{40}{40 + e ^{x}} d x

x - ln ∣ e^{x} + 40 ∣ + C

Solution steps

\int \frac{40}{40 + e ^{x}} d x

Take the constant out:

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

= 40 \cdot \int \frac{1}{40 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 40 + u )} d u

= 40 \cdot \int \frac{1}{u ( 40 + u )} d u

Take the partial fraction of

\frac{1}{u ( 40 + u )} : \frac{1}{40 u} - \frac{1}{40 ( u + 40 )}

= 40 \cdot \int \frac{1}{40 u} - \frac{1}{40 ( u + 40 )} 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

= 40 (\int \frac{1}{40 u} d u - \int \frac{1}{40 ( u + 40 )} d u)

\int \frac{1}{40 u} d u = \frac{1}{40} ln ∣ u ∣

\int \frac{1}{40 ( u + 40 )} d u = \frac{1}{40} ln ∣ u + 40 ∣

= 40 (\frac{1}{40} ln ∣ u ∣ - \frac{1}{40} ln ∣ u + 40 ∣)

Substitute back

u = e^{x}

= 40 (\frac{1}{40} ln ∣ e^{x} ∣ - \frac{1}{40} ln ∣ e^{x} + 40 ∣)

Simplify

40 (\frac{1}{40} ln ∣ e^{x} ∣ - \frac{1}{40} ln ∣ e^{x} + 40 ∣) : x - ln ∣ e^{x} + 40 ∣

= x - ln ∣ e^{x} + 40 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 40 ∣ + C

f (x) = 2 e^{x - 1}

\int \frac{9}{x + 1} d x

\frac{d}{d x} (3 (- csc^{2} (x) + sec^{2} (x)))

y = 7 x^{2}, y^{2} = \frac{1}{7} x

\frac{1}{1 + x}

What is the integral of (40)/(40+e^x) ?
The integral of (40)/(40+e^x) is x-ln|e^x+40|+C