To compute the integral [tex]\(\int \frac{x \, dx}{x^2 + 4x - 12}\)[/tex], let us go through the steps:
1. Factoring the Denominator:
First, we factor the quadratic expression in the denominator [tex]\(x^2 + 4x - 12\)[/tex].
We look for two numbers that multiply to [tex]\(-12\)[/tex] (the constant term) and add to [tex]\(4\)[/tex] (the coefficient of [tex]\(x\)[/tex]). Those numbers are [tex]\(6\)[/tex] and [tex]\(-2\)[/tex]. Thus,
[tex]\[
x^2 + 4x - 12 = (x + 6)(x - 2)
\][/tex]
2. Substitute the Factored Denominator:
We rewrite the integral with the factored denominator:
[tex]\[
\int \frac{x \, dx}{(x + 6)(x - 2)}
\][/tex]
3. Partial Fraction Decomposition:
We decompose the integrand into partial fractions. Assume:
[tex]\[
\frac{x}{(x + 6)(x - 2)} = \frac{A}{x + 6} + \frac{B}{x - 2}
\][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we multiply both sides by [tex]\((x + 6)(x - 2)\)[/tex]:
[tex]\[
x = A(x - 2) + B(x + 6)
\][/tex]
We obtain the equation by equating coefficients:
[tex]\[
x = Ax - 2A + Bx + 6B
\][/tex]
Combining like terms:
[tex]\[
x = (A + B)x + (-2A + 6B)
\][/tex]
Equating coefficients of [tex]\(x\)[/tex] and the constant term, we get:
- Coefficient of [tex]\(x\)[/tex]: [tex]\(A + B = 1\)[/tex]
- Constant term: [tex]\( -2A + 6B = 0\)[/tex]
Solving this system of equations:
[tex]\[
A + B = 1 \quad \text{(i)}
\][/tex]
[tex]\[
-2A + 6B = 0 \quad \text{(ii)}
\][/tex]
From (ii):
[tex]\[
-2A + 6B = 0 \implies B = \frac{A}{3}
\][/tex]
Substitute [tex]\(B = \frac{A}{3}\)[/tex] into (i):
[tex]\[
A + \frac{A}{3} = 1 \implies \frac{3A + A}{3} = 1 \implies \frac{4A}{3} = 1 \implies 4A = 3 \implies A = \frac{3}{4}
\][/tex]
Thus,
[tex]\[
B = \frac{A}{3} = \frac{3/4}{3} = \frac{1}{4}
\][/tex]
Hence, we decompose the fraction:
[tex]\[
\frac{x}{(x + 6)(x - 2)} = \frac{3/4}{x + 6} + \frac{1/4}{x - 2}
\][/tex]
4. Integrating Each Term:
Now, we integrate each partial fraction term separately:
[tex]\[
\int \frac{x \, dx}{(x + 6)(x - 2)} = \int \left( \frac{3/4}{x + 6} + \frac{1/4}{x - 2} \right) dx
\][/tex]
This becomes:
[tex]\[
\int \frac{3/4}{x + 6} \, dx + \int \frac{1/4}{x - 2} \, dx
\][/tex]
Using the integral formula [tex]\(\int \frac{1}{u} \, du = \ln|u|\)[/tex], these integrals evaluate to:
[tex]\[
\frac{3}{4} \int \frac{1}{x + 6} \, dx + \frac{1}{4} \int \frac{1}{x - 2} \, dx
\][/tex]
[tex]\[
= \frac{3}{4} \ln|x + 6| + \frac{1}{4} \ln|x - 2| + C
\][/tex]
Combining the logarithms:
[tex]\[
= \frac{1}{4} \ln|x - 2| + \frac{3}{4} \ln|x + 6| + C
\][/tex]
Thus, the integral [tex]\(\int \frac{x \, dx}{x^2 + 4x - 12}\)[/tex] is:
[tex]\[
\boxed{\frac{1}{4} \ln|x - 2| + \frac{3}{4} \ln|x + 6| + C}
\][/tex]
