What is the integral of 1/(e^x-3) ?

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

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integral of 1/(e^x-3)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

= \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

= - \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 ∣

= - \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣

Substitute back

u = e^{x}

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (cos (x^{2} + 1)))

f (v) = e^{2 s i n (3 v)}

\int sin (x) - cos (x) d x

n = 1 \sum \infty (- 3)^{n}

\frac{d y}{d x} = \frac{x - 3}{y + 5}

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