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

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

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

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

- \frac{ln 2 \cdot 3 ^{x - 1} + \frac{2}{3} - ln 2 \cdot 3 ^{x - 1} - \frac{4}{3}}{3 ln ( 3 )} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot \int \frac{1}{u ^{2} - u - 2} d u

Complete the square

u^{2} - u - 2 : (u - \frac{1}{2})^{2} - \frac{9}{4}

= \frac{1}{ln ( 3 )} \cdot \int \frac{1}{( u - \frac{1}{2} ) ^{2} - \frac{9}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} - 9} d v

= \frac{1}{ln ( 3 )} \cdot \int \frac{4}{4 v ^{2} - 9} d v

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \int \frac{1}{4 v ^{2} - 9} d v

Apply Integral Substitution

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \int \frac{1}{6 ( w ^{2} - 1 )} d w

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \frac{1}{6} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

w^{2} - 1 : - (- w^{2} + 1)

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \frac{1}{6} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \frac{1}{6} (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \frac{1}{6} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= \frac{1}{ln ( 3 )} \cdot 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( 3 ^{x} - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( 3 ^{x} - \frac{1}{2} ) - 1}{2}))

Simplify

\frac{1}{ln ( 3 )} \cdot 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( 3 ^{x} - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( 3 ^{x} - \frac{1}{2} ) - 1}{2})) : - \frac{ln 2 \cdot 3 ^{x - 1} + \frac{2}{3} - ln 2 \cdot 3 ^{x - 1} - \frac{4}{3}}{3 ln ( 3 )}

= - \frac{ln 2 \cdot 3 ^{x - 1} + \frac{2}{3} - ln 2 \cdot 3 ^{x - 1} - \frac{4}{3}}{3 ln ( 3 )}

Add a constant to the solution

= - \frac{ln 2 \cdot 3 ^{x - 1} + \frac{2}{3} - ln 2 \cdot 3 ^{x - 1} - \frac{4}{3}}{3 ln ( 3 )} + C

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