What is the integral of 4/(3-e^x) ?

The integral of 4/(3-e^x) is 4(1/3 x-1/3 ln|e^x-3|)+C

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integral of 4/(3-e^x)

\int \frac{4}{3 - e ^{x}} d x

4 (\frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} - 3 ∣) + C

Solution steps

\int \frac{4}{3 - e ^{x}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{3 - e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 3 - u )} d u

= 4 \cdot \int \frac{1}{u ( 3 - u )} d u

Take the partial fraction of

\frac{1}{u ( 3 - u )} : \frac{1}{3 u} - \frac{1}{3 ( u - 3 )}

= 4 \cdot \int \frac{1}{3 u} - \frac{1}{3 ( u - 3 )} 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

= 4 (\int \frac{1}{3 u} d u - \int \frac{1}{3 ( u - 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u - 3 )} d u = \frac{1}{3} ln ∣ u - 3 ∣

= 4 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u - 3 ∣)

Substitute back

u = e^{x}

= 4 (\frac{1}{3} ln ∣ e^{x} ∣ - \frac{1}{3} ln ∣ e^{x} - 3 ∣)

\frac{1}{3} ln ∣ e^{x} ∣ = \frac{1}{3} x

= 4 (\frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} - 3 ∣)

Add a constant to the solution

= 4 (\frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} - 3 ∣) + C

x^{2} + 2 x + 11, - 4 x + 2, [0]

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

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x^{2} y^{2})

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

What is the integral of 4/(3-e^x) ?
The integral of 4/(3-e^x) is 4(1/3 x-1/3 ln|e^x-3|)+C