Question

Use partial fractions to find the indefinite integral. $$ \int \frac{-2}{x^{2}-16} d x $$

Short answer

The integral of \(-\frac{2}{x^{2}-16}\) with respect to x is \(-\frac {1}{4} \ln |x-4| + \frac {1}{4} \ln |x+4| + C\).

Step-by-step solution

Decompose into Partial Fractions

The first task is to decompose the given fraction into partial fractions. The process involves expressing a given fraction as the sum of simpler fractions. Here, it's also important to notice that \(x^{2}-16\) is a difference of squares: \((x-4)(x+4)\), so this helps simplify the problem. Thus, \(-\frac{2}{x^{2}-16}\) can be rewritten as \(\frac{A}{x-4} + \frac{B}{x+4}\), where A and B are constants that we need to solve for.

Solve for Constants A and B

Next, solve for A and B by multiplying both sides of the equation by the denominator on the left side, which results in the equation: \(-2 = A(x+4) + B(x-4)\).Setting \(x = 4\) gives \(A \cdot 8 = -2\), hence \(A = -\frac{1}{4}\).Setting \(x = -4\) gives \(B \cdot -8 = -2\), hence \(B = \frac{1}{4}\).

Replace Constants in the Expression

Replace the values of A and B into the expression. We therefore get \(-\frac{1}{4x-16} + \frac{1}{4x+16}\).

Integrate each Term Separately

The integral of the sum of functions is equal to the sum of the integrals of the functions. This allows the integrand to be split into separate fractions which can be integrated separately. Thus, \(\int \frac{-2}{x^{2}-16} dx = \int \frac{A}{x-4} dx + \int \frac{B}{x+4} dx\).

Evaluate the Integral

Evaluate each integral separately. The integral of \(1/x\) with respect to x is \(\ln |x|\), applying this we get \(-\frac {1}{4}\int\frac {1}{x-4} dx+ \frac {1}{4}\int\frac {1}{x+4} dx\), which equals to \(-\frac {1}{4} \ln |x-4| + \frac {1}{4} \ln |x+4| + C\), where C is the constant of integration.