What is the integral of (10)/(10+e^x) ?

The integral of (10)/(10+e^x) is x-ln|e^x+10|+C

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integral of (10)/(10+e^x)

\int \frac{10}{10 + e ^{x}} d x

x - ln ∣ e^{x} + 10 ∣ + C

Solution steps

\int \frac{10}{10 + e ^{x}} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{10 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 10 + u )} d u

= 10 \cdot \int \frac{1}{u ( 10 + u )} d u

Take the partial fraction of

\frac{1}{u ( 10 + u )} : \frac{1}{10 u} - \frac{1}{10 ( u + 10 )}

= 10 \cdot \int \frac{1}{10 u} - \frac{1}{10 ( u + 10 )} 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

= 10 (\int \frac{1}{10 u} d u - \int \frac{1}{10 ( u + 10 )} d u)

\int \frac{1}{10 u} d u = \frac{1}{10} ln ∣ u ∣

\int \frac{1}{10 ( u + 10 )} d u = \frac{1}{10} ln ∣ u + 10 ∣

= 10 (\frac{1}{10} ln ∣ u ∣ - \frac{1}{10} ln ∣ u + 10 ∣)

Substitute back

u = e^{x}

= 10 (\frac{1}{10} ln ∣ e^{x} ∣ - \frac{1}{10} ln ∣ e^{x} + 10 ∣)

Simplify

10 (\frac{1}{10} ln ∣ e^{x} ∣ - \frac{1}{10} ln ∣ e^{x} + 10 ∣) : x - ln ∣ e^{x} + 10 ∣

= x - ln ∣ e^{x} + 10 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 10 ∣ + C

\frac{d}{d x} (\frac{( x + y )}{( x - y )})

x \to 0 lim (- 3 - 2 x)

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

\frac{d y}{d x} = x^{2} e^{x^{3}} y^{3}

\frac{\partial}{\partial x} (sin (π (3 x - 2 y)))

What is the integral of (10)/(10+e^x) ?
The integral of (10)/(10+e^x) is x-ln|e^x+10|+C