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Chapter 12: Problem 21

Use partial fractions to find the indefinite integral. $$ \int \frac{3}{x^{2}+x-2} d x $$

Short Answer

Expert verified
The solution to \(\int \frac{3}{x^{2}+x-2} dx\) is \(ln |x - 1| - ln |x+2| + C\).

Step by step solution

01

Factor the quadratic

Firstly, it's important to factor the denominator in the integral. To do this, find two numbers that multiplied are equal to -2 (c in the quadratic equation), and added are equal to 1 (b in the quadratic equation). The equation \(x^2+x-2\) factors into \((x-1)(x+2)\). So, we rewrite the integral: \[\int \frac{3}{(x-1)(x+2)} dx \]
02

Decomposition into Partial Fractions

Now, we need to decompose the fraction into partial fractions. We write \[\frac{3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\]Cross-multiplying gives:\[3 = A(x + 2) + B(x - 1)\]Setting \(x = 1\), we get \(A=1\). Setting \(x = -2\), we find \(B=1\). Therefore, we may write the integral: \[\int \frac{1}{x-1} dx + \int \frac{1}{x+2} dx \]
03

Integral computation

The integrals we have are now straightforward to compute. The antiderivative of \(\frac{1}{x-a}\) is \(ln |x-a|\). So the integral of \(\frac{1}{x-1}\) is \(ln |x - 1|\) and the integral of \(\frac{1}{x+2}\) is \(ln |x+2|\). Thus the integral of interest is: \[\int \frac{1}{x-1} dx + \int \frac{1}{x+2} dx = ln |x - 1| - ln |x+2| + C \]where \(C\) is the constant of integration.

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Most popular questions from this chapter

Use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the indicated value of \(n\). Compare these results with the exact value of the definite integral. Round your answers to four decimal places. $$ \int_{0}^{2}\left(x^{4}+1\right) d x, n=4 $$

Present Value In Exercises 25 and 26, use a program similar to the Simpson's Rule program on page 906 with \(n=8\) to approximate the present value of the income \(c(t)\) over \(t_{1}\) years at the given annual interest rate \(r\). Then use the integration capabilities of a graphing utility to approximate the present value. Compare the results. (Present value is defined in Section 12.1.) $$ c(t)=200,000+15,000 \sqrt[3]{t}, r=10 \%, t_{1}=8 $$

Revenue The revenue (in dollars per year) for a new product is modeled by \(R=10,000\left[1-\frac{1}{\left(1+0.1 t^{2}\right)^{1 / 2}}\right]\) where \(t\) is the time in years. Estimate the total revenue from sales of the product over its first 2 years on the market.

Present Value In Exercises 25 and 26, use a program similar to the Simpson's Rule program on page 906 with \(n=8\) to approximate the present value of the income \(c(t)\) over \(t_{1}\) years at the given annual interest rate \(r\). Then use the integration capabilities of a graphing utility to approximate the present value. Compare the results. (Present value is defined in Section 12.1.) $$ c(t)=6000+200 \sqrt{t}, r=7 \%, t_{1}=4 $$

Use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the indicated value of \(n\). Compare these results with the exact value of the definite integral. Round your answers to four decimal places. $$ \int_{0}^{2} x^{3} d x, n=8 $$
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