How do you integrate #8/((x-3)^2+1)# using partial fractions?

Question

Luke Alvarez · Accepted Answer

See the answer below:

William Abdalla · Answer

undefined To integrate $ \frac{8}{(x-3)^2 + 1} $ using partial fractions, we need to express the fraction as a sum of simpler fractions.

First, let's express the denominator as a sum of two terms using the following form:
$$ (x-3)^2 + 1 = (x-3)^2 + 1^2 $$

We can rewrite this as:
$$ (x-3)^2 + 1 = (x-3 + i)(x-3 - i) $$

Thus, we can express $ \frac{8}{(x-3)^2 + 1} $ as:
$$ \frac{8}{(x-3)^2 + 1} = \frac{A}{x-3+i} + \frac{B}{x-3-i} $$

Now, we need to find the values of $ A $ and $ B $ by equating the numerators:
$$ 8 = A(x-3-i) + B(x-3+i) $$

Expanding and combining like terms:
$$ 8 = Ax - 3A - Ai + Bx - 3B + Bi $$

Grouping the terms with $ x $ and the constants:
$$ 8 = (A + B)x + (-3A - 3B) + (B - A)i $$

From the above equation, equating the real and imaginary parts separately:
Real part:
$$ 8 = (A + B)x - 3(A + B) $$
$$ 8 = (A + B)(x - 3) $$

Imaginary part:
$$ 0 = (B - A)i $$

From the imaginary part, we get:
$$ B = A $$

Substituting this into the real part equation:
$$ 8 = 2Ax - 6A $$

Solving for $ A $:
$$ A = \frac{4}{x-3+i} $$

Since $ B = A $, $ B = \frac{4}{x-3-i} $.

Now, integrating each term:
$$ \int \frac{4}{x-3+i} \, dx = 4\ln|x-3+i| + C_1 $$
$$ \int \frac{4}{x-3-i} \, dx = 4\ln|x-3-i| + C_2 $$

Finally, combining the results:
$$ \int \frac{8}{(x-3)^2 + 1} \, dx = 4\ln|x-3+i| + 4\ln|x-3-i| + C $$
$$ = 4\ln|(x-3+i)(x-3-i)| + C $$
$$ = 4\ln|(x-3)^2 + 1| + C $$