What is the integral of (31)/(31+e^x) ?

The integral of (31)/(31+e^x) is x-ln|e^x+31|+C

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integral of (31)/(31+e^x)

\int \frac{31}{31 + e ^{x}} d x

x - ln ∣ e^{x} + 31 ∣ + C

Solution steps

\int \frac{31}{31 + e ^{x}} d x

Take the constant out:

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

= 31 \cdot \int \frac{1}{31 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 31 + u )} d u

= 31 \cdot \int \frac{1}{u ( 31 + u )} d u

Take the partial fraction of

\frac{1}{u ( 31 + u )} : \frac{1}{31 u} - \frac{1}{31 ( u + 31 )}

= 31 \cdot \int \frac{1}{31 u} - \frac{1}{31 ( u + 31 )} 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

= 31 (\int \frac{1}{31 u} d u - \int \frac{1}{31 ( u + 31 )} d u)

\int \frac{1}{31 u} d u = \frac{1}{31} ln ∣ u ∣

\int \frac{1}{31 ( u + 31 )} d u = \frac{1}{31} ln ∣ u + 31 ∣

= 31 (\frac{1}{31} ln ∣ u ∣ - \frac{1}{31} ln ∣ u + 31 ∣)

Substitute back

u = e^{x}

= 31 (\frac{1}{31} ln ∣ e^{x} ∣ - \frac{1}{31} ln ∣ e^{x} + 31 ∣)

Simplify

31 (\frac{1}{31} ln ∣ e^{x} ∣ - \frac{1}{31} ln ∣ e^{x} + 31 ∣) : x - ln ∣ e^{x} + 31 ∣

= x - ln ∣ e^{x} + 31 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 31 ∣ + C

\int (3 x + 1) (3 x^{2} + 2 x)^{\frac{3}{4}} d x

y = 4 x^{4} + 3 x - 5, (1, 2)

f (x) = \frac{1}{2 + 3 x}, at x = 1

y = - 3 x - 5

\frac{\partial}{\partial x} (- 2 (y + 1) x^{- 3})

What is the integral of (31)/(31+e^x) ?
The integral of (31)/(31+e^x) is x-ln|e^x+31|+C