How do you integrate #int [ e^x / ((e^x -4) (e^x + 6)) ] dx# using partial fractions?

Question

Elijah Abram · Accepted Answer

See the answer below:

Continuation:

Isabella Ackerman · Answer

undefined To integrate $ \frac{e^x}{(e^x - 4)(e^x + 6)} $ using partial fractions, you can first express the integrand as a sum of simpler fractions:

$$ \frac{e^x}{(e^x - 4)(e^x + 6)} = \frac{A}{e^x - 4} + \frac{B}{e^x + 6} $$

Multiply both sides by $ (e^x - 4)(e^x + 6) $ to clear the denominators:

$$ e^x = A(e^x + 6) + B(e^x - 4) $$

Expand and group like terms:

$$ e^x = (A + B)e^x + 6A - 4B $$

Compare coefficients of $ e^x $ on both sides:

$$ 1 = A + B $$
$$ 0 = 6A - 4B $$

Solve the system of equations for $ A $ and $ B $:

$$ A = \frac{2}{5} $$
$$ B = \frac{3}{5} $$

Now that you have the values of $ A $ and $ B $, rewrite the integrand using partial fractions:

$$ \frac{e^x}{(e^x - 4)(e^x + 6)} = \frac{\frac{2}{5}}{e^x - 4} + \frac{\frac{3}{5}}{e^x + 6} $$

Integrate each term separately:

$$ \int \frac{e^x}{(e^x - 4)(e^x + 6)} \, dx = \frac{2}{5} \int \frac{1}{e^x - 4} \, dx + \frac{3}{5} \int \frac{1}{e^x + 6} \, dx $$

$$ = \frac{2}{5} \ln|e^x - 4| + \frac{3}{5} \ln|e^x + 6| + C $$

Where $ C $ is the constant of integration.