What is the integral of 6/(6+e^x) ?

The integral of 6/(6+e^x) is x-ln|e^x+6|+C

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integral of 6/(6+e^x)

\int \frac{6}{6 + e ^{x}} d x

x - ln ∣ e^{x} + 6 ∣ + C

Solution steps

\int \frac{6}{6 + e ^{x}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{6 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 6 + u )} d u

= 6 \cdot \int \frac{1}{u ( 6 + u )} d u

Take the partial fraction of

\frac{1}{u ( 6 + u )} : \frac{1}{6 u} - \frac{1}{6 ( u + 6 )}

= 6 \cdot \int \frac{1}{6 u} - \frac{1}{6 ( u + 6 )} 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

= 6 (\int \frac{1}{6 u} d u - \int \frac{1}{6 ( u + 6 )} d u)

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

\int \frac{1}{6 ( u + 6 )} d u = \frac{1}{6} ln ∣ u + 6 ∣

= 6 (\frac{1}{6} ln ∣ u ∣ - \frac{1}{6} ln ∣ u + 6 ∣)

Substitute back

u = e^{x}

= 6 (\frac{1}{6} ln ∣ e^{x} ∣ - \frac{1}{6} ln ∣ e^{x} + 6 ∣)

Simplify

6 (\frac{1}{6} ln ∣ e^{x} ∣ - \frac{1}{6} ln ∣ e^{x} + 6 ∣) : x - ln ∣ e^{x} + 6 ∣

= x - ln ∣ e^{x} + 6 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 6 ∣ + C

\int \frac{a + x}{a - x} d x

\int \frac{1}{e ^{2 t} - 2} d t

\frac{d}{d x} (0.6 x^{3} 4^{x})

y = 3 x^{2}, y = 2, x = 0, x \geq 0

\frac{d}{d x} (x e^{- \frac{x ^{2}}{2}})