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# How do I evaluate intdx/(e^(x)(3e^(x)+2))?

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#### Explanation

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#### Explanation:

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1
Jun 18, 2018

The answer is $= - \frac{1}{2 {e}^{x}} - \frac{3}{4} x + \frac{3}{4} \ln \left(3 {e}^{x} + 2\right) + C$

#### Explanation:

Let ${e}^{x} = u$, $\implies$, $\mathrm{du} = {e}^{x} \mathrm{dx}$

Therefore, the integral is

$I = \int \frac{\mathrm{dx}}{{e}^{x} \left(3 {e}^{x} + 2\right)} = \int \frac{\mathrm{du}}{{e}^{2 x} \left(3 {e}^{x} + 2\right)}$

$= \int \frac{\mathrm{du}}{{u}^{2} \left(3 u + 2\right)}$

Perform the decomposition into partial fractions

$\frac{1}{{u}^{2} \left(3 u + 2\right)} = \frac{A}{u} ^ 2 + \frac{B}{u} + \frac{C}{3 u + 2}$

$= \frac{A \left(3 u + 2\right) + B u \left(3 u + 2\right) + C {u}^{2}}{{u}^{2} \left(3 u + 2\right)}$

The denominator is the same, compare the numerators

$1 = A \left(3 u + 2\right) + B u \left(3 u + 2\right) + C {u}^{2}$

Let $u = 0$, $\implies$, $1 = 2 A$, $\implies$, $A = \frac{1}{2}$

Coefficients of $u$

$0 = 3 A + 2 B$, $\implies$, $B = - \frac{3}{2} A = - \frac{3}{4}$

Coefficients of ${u}^{2}$

$0 = 3 B + C$, $\implies$, $C = - 3 B = \frac{9}{4}$

Therefore,

$\frac{1}{{u}^{2} \left(3 u + 2\right)} = \frac{\frac{1}{2}}{u} ^ 2 + \frac{- \frac{3}{4}}{u} + \frac{\frac{9}{4}}{3 u + 2}$

So,

$\int \frac{\mathrm{du}}{{u}^{2} \left(3 u + 2\right)} = \frac{1}{2} \int \frac{\mathrm{du}}{u} ^ 2 - \frac{3}{4} \int \frac{\mathrm{du}}{u} + \frac{9}{4} \int \frac{\mathrm{du}}{3 u + 2}$

$= - \frac{1}{2 u} - \frac{3}{4} \ln u + \frac{3}{4} \ln \left(3 u + 2\right)$

$= - \frac{1}{2 {e}^{x}} - \frac{3}{4} \ln \left({e}^{x}\right) + \frac{3}{4} \ln \left(3 {e}^{x} + 2\right) + C$

$= - \frac{1}{2 {e}^{x}} - \frac{3}{4} x + \frac{3}{4} \ln \left(3 {e}^{x} + 2\right) + C$

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