What is the integral of (e^x)/((e^x-3)(e^x+2)) ?

The integral of (e^x)/((e^x-3)(e^x+2)) is 1/5 ln|e^x-3|-1/5 ln|e^x+2|+C

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integral of (e^x)/((e^x-3)(e^x+2))

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

\frac{1}{5} ln ∣ e^{x} - 3 ∣ - \frac{1}{5} ln ∣ e^{x} + 2 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{( u - 3 ) ( u + 2 )} d u

= \int \frac{1}{( u - 3 ) ( u + 2 )} d u

Take the partial fraction of

\frac{1}{( u - 3 ) ( u + 2 )} : \frac{1}{5 ( u - 3 )} - \frac{1}{5 ( u + 2 )}

= \int \frac{1}{5 ( u - 3 )} - \frac{1}{5 ( u + 2 )} 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}{5 ( u - 3 )} d u - \int \frac{1}{5 ( u + 2 )} d u

\int \frac{1}{5 ( u - 3 )} d u = \frac{1}{5} ln ∣ u - 3 ∣

\int \frac{1}{5 ( u + 2 )} d u = \frac{1}{5} ln ∣ u + 2 ∣

= \frac{1}{5} ln ∣ u - 3 ∣ - \frac{1}{5} ln ∣ u + 2 ∣

Substitute back

u = e^{x}

= \frac{1}{5} ln ∣ e^{x} - 3 ∣ - \frac{1}{5} ln ∣ e^{x} + 2 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ e^{x} - 3 ∣ - \frac{1}{5} ln ∣ e^{x} + 2 ∣ + C

f (x) = csc^{7} (2 x^{π} + π^{3})

y = x^{2} - 7 x + 7, (4, 0)

y^{^{''}} - 9 y = 0

\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 2}{x})

\frac{d y}{d x} = \frac{- y ^{2}}{1 + y ^{2}} e^{- x}

What is the integral of (e^x)/((e^x-3)(e^x+2)) ?
The integral of (e^x)/((e^x-3)(e^x+2)) is 1/5 ln|e^x-3|-1/5 ln|e^x+2|+C