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

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

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

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

2 (\frac{1}{4} ln - 4 e^{x} + 4 e^{2 x} - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} e^{x} - ln - \frac{6}{5} + \frac{2}{5} e^{x})) + C

Solution steps

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

Apply u-substitution

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

Complete the square

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

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

\frac{2 v + 1}{4 v ^{2} - 25} : \frac{2 v}{4 v ^{2} - 25} + \frac{1}{4 v ^{2} - 25}

Apply the Sum Rule:

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

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

\int \frac{2 v}{4 v ^{2} - 25} d v = \frac{1}{4} ln 4 v^{2} - 25

\int \frac{1}{4 v ^{2} - 25} d v = - \frac{1}{20} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1)

= 2 (\frac{1}{4} ln 4 v^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1))

Substitute back

= 2 (\frac{1}{4} ln 4 (e^{x} - \frac{1}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (e^{x} - \frac{1}{2}) + 1 - ln \frac{2}{5} (e^{x} - \frac{1}{2}) - 1))

Simplify

2 (\frac{1}{4} ln 4 (e^{x} - \frac{1}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (e^{x} - \frac{1}{2}) + 1 - ln \frac{2}{5} (e^{x} - \frac{1}{2}) - 1)) : 2 (\frac{1}{4} ln - 4 e^{x} + 4 e^{2 x} - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} e^{x} - ln - \frac{6}{5} + \frac{2}{5} e^{x}))

= 2 (\frac{1}{4} ln - 4 e^{x} + 4 e^{2 x} - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} e^{x} - ln - \frac{6}{5} + \frac{2}{5} e^{x}))

Add a constant to the solution

= 2 (\frac{1}{4} ln - 4 e^{x} + 4 e^{2 x} - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} e^{x} - ln - \frac{6}{5} + \frac{2}{5} e^{x})) + C

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