What is the integral of (e^{3x})/(e^{2x)-e^x-6} ?

The integral of (e^{3x})/(e^{2x)-e^x-6} is e^x-1/2-4/5 ln|2e^x+4|+9/5 ln|2e^x-6|+C

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integral of (e^{3x})/(e^{2x)-e^x-6}

\int \frac{e ^{3 x}}{e ^{2 x} - e ^{x} - 6} d x

e^{x} - \frac{1}{2} - \frac{4}{5} ln ∣ 2 e^{x} + 4 ∣ + \frac{9}{5} ln ∣ 2 e^{x} - 6 ∣ + C

Solution steps

\int \frac{e ^{3 x}}{e ^{2 x} - e ^{x} - 6} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} - u - 6} d u

= \int \frac{u ^{2}}{u ^{2} - u - 6} d u

Complete the square

u^{2} - u - 6 : (u - \frac{1}{2})^{2} - \frac{25}{4}

= \int \frac{u ^{2}}{( u - \frac{1}{2} ) ^{2} - \frac{25}{4}} d u

Apply u-substitution

: \int \frac{( 2 v + 1 ) ^{2}}{4 v ^{2} - 25} d v

= \int \frac{( 2 v + 1 ) ^{2}}{4 v ^{2} - 25} d v

Take the partial fraction of

\frac{( 2 v + 1 ) ^{2}}{4 v ^{2} - 25} : 1 - \frac{8}{5 ( 2 v + 5 )} + \frac{18}{5 ( 2 v - 5 )}

= \int 1 - \frac{8}{5 ( 2 v + 5 )} + \frac{18}{5 ( 2 v - 5 )} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d v - \int \frac{8}{5 ( 2 v + 5 )} d v + \int \frac{18}{5 ( 2 v - 5 )} d v

\int 1 d v = v

\int \frac{8}{5 ( 2 v + 5 )} d v = \frac{4}{5} ln ∣ 2 v + 5 ∣

\int \frac{18}{5 ( 2 v - 5 )} d v = \frac{9}{5} ln ∣ 2 v - 5 ∣

= v - \frac{4}{5} ln ∣ 2 v + 5 ∣ + \frac{9}{5} ln ∣ 2 v - 5 ∣

Substitute back

= e^{x} - \frac{1}{2} - \frac{4}{5} ln 2 (e^{x} - \frac{1}{2}) + 5 + \frac{9}{5} ln 2 (e^{x} - \frac{1}{2}) - 5

Simplify

e^{x} - \frac{1}{2} - \frac{4}{5} ln 2 (e^{x} - \frac{1}{2}) + 5 + \frac{9}{5} ln 2 (e^{x} - \frac{1}{2}) - 5 : e^{x} - \frac{1}{2} - \frac{4}{5} ln ∣ 2 e^{x} + 4 ∣ + \frac{9}{5} ln ∣ 2 e^{x} - 6 ∣

= e^{x} - \frac{1}{2} - \frac{4}{5} ln ∣ 2 e^{x} + 4 ∣ + \frac{9}{5} ln ∣ 2 e^{x} - 6 ∣

Add a constant to the solution

= e^{x} - \frac{1}{2} - \frac{4}{5} ln ∣ 2 e^{x} + 4 ∣ + \frac{9}{5} ln ∣ 2 e^{x} - 6 ∣ + C

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What is the integral of (e^{3x})/(e^{2x)-e^x-6} ?
The integral of (e^{3x})/(e^{2x)-e^x-6} is e^x-1/2-4/5 ln|2e^x+4|+9/5 ln|2e^x-6|+C