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Chapter 12: Problem 15

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

Short Answer

Expert verified
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

01

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.
02

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 \(B = \frac{1}{4}\).
03

Replace Constants in the Expression

Replace the values of A and B into the expression. We therefore get \(-\frac{1}{4x-16} + \frac{1}{4x+16}\).
04

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, \(\int \frac{-2}{x^{2}-16} dx = \int \frac{A}{x-4} dx + \int \frac{B}{x+4} dx\).
05

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.

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Most popular questions from this chapter

Decide whether the integral is improper. Explain your reasoning. $$ \int_{0}^{1} \frac{d x}{3 x-2} $$

Use the error formulas to find \(n\) such that the error in the approximation of the definite integral is less than \(0.0001\) using (a) the Trapezoidal Rule and (b) Simpson's Rule. $$ \int_{0}^{1} x^{3} d x $$

Use a spreadsheet to complete the table for the specified values of \(a\) and \(n\) to demonstrate that \(\lim _{x \rightarrow \infty} x^{n} e^{-a x}=0, \quad a>0, n>0\) \begin{tabular}{|l|l|l|l|l|} \hline\(x\) & 1 & 10 & 25 & 50 \\ \hline\(x^{n} e^{-a x}\) & & & & \\ \hline \end{tabular} $$ a=1, n=1 $$

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{27} \frac{5}{\sqrt[3]{x}} d x $$

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n\). (Round your answers to three significant digits.) $$ \int_{0}^{2} \frac{1}{\sqrt{1+x^{3}}} d x, n=4 $$
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