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Chapter 5: Q. 13 (page 441)

In Exercises 12–14, suppose that you want to obtain a partialfraction decomposition of a rational function $\frac{p (x)}{q (x)}$ according to Theorem 5.12.
If q(x) is an irreducible quadratic, what can you say about its partial-fraction decomposition? What if q(x) is a reducible quadratic? Consider the functions $q (x) = x^{2} + 1 a n d q (x) = (x - 1) (x - 2)$ to find your answer.

Short Answer

Expert verified

If the function is irreducible, nothing further can be done but if function is reducible we get, $\frac{A_{1}}{l_{1} (x)} + \frac{A_{2}}{l_{2} (x)} where l_{1} (x) = x - 1 and l_{2} (x) = x - 2$

Step by step solution

01

Step 1. Given Information

The given functions are $q (x) = x^{2} + 1 a n d q (x) = (x - 1) (x - 2)$

02

Step 2. Explanation

If q(x) is an irreducible quadratic then $\frac{p (x)}{q (x)}$ is its own partial fractions decomposition, there is nothing further we can decompose.

If q(x) is a reducible quadratic then q(x) is a product of $l_{1} (x) and l_{2} (x)$ of two linear functions and obtain a partial decomposition of the form

$\frac{A_{1}}{l_{1} (x)} + \frac{A_{2}}{l_{2} (x)}$

As the given function is

$q (x) = x^{2} + 1, So, l_{1} (x) = x - 1 and l_{2} (x) = x - 2$

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Most popular questions from this chapter

Why don’t we need to have a square root involved in order to apply trigonometric substitution with the tangent? In other words, why can we use the substitution $x = a \tan u$ when we see $x^{2} + a^{2}$ , even though we can’t use the substitution $x = a \sin u$ unless the integrand involves the square root of $a^{2} - x^{2}$ ? (Hint: Think about domains.)

For each function u(x) in Exercises 9–12, write the differential du in terms of the differential dx.

$u (x) = x^{2} + 1$

Why don’t we ever have cause to use the trigonometric substitution $x = \cos u$ ?

Solve given definite integral.

$\int_{- 1}^{1} x^{3} \sqrt{9 x^{2} - 1} d x$

Solve the integral: $\int x \sin {(x)}^{2} d x$ .

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