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

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

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integral of (e^x)/(e^x(e^x-3))

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

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

Solution steps

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

Cancel the common factor:

e^{x}

= \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{( 2 x ) ^{3}}{5}

\frac{d y}{d x} (y + 4) = 2 x y + 4 y

(x y^{^{'}} - y) cos (\frac{y}{x}) + x = 0

\frac{d}{d x} (cos (\frac{x ^{3}}{10}))

\frac{d y}{d x} = \frac{( x - 3 ) ^{2}}{x}

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