Evaluate the integral. ∫(10 x+2)/(x^3-5 x^2+x-5) d x ; use (A)/(x-5)+(B x+C)/(x^2+1)

Name: Problem 17, Chapter 7: Calculus: Single Variable
Uploaded: 2022-02-11T17:33:48-08:00
Duration: 7 min 3 s
Channel: Noah Fisher
Description: Evaluate the integral. ∫(10 x+2)/(x^3-5 x^2+x-5) d x ; use (A)/(x-5)+(B x+C)/(x^2+1)

Evaluate the integral. ∫(10 x+2)/(x^3-5 x^2+x-5) d x ; use ...

Key Concepts

Factorization of Polynomials

This concept involves expressing a given polynomial as a product of simpler polynomials. In the context of integration, especially rational functions, factorizing the denominator into irreducible factors (such as a linear and an irreducible quadratic factor) is essential to set up a partial fractions decomposition. It simplifies the integral by breaking it into parts that are easier to integrate.

Partial Fractions Decomposition

This technique is used to break down a complex rational function into a sum of simpler fractions, each with a simpler denominator. By expressing the integrand in the form of A/(x - 5) + (Bx + C)/(x^2 + 1), the integral can be split into standard integrals which are then more straightforward to evaluate.

Integration of Rational Functions

After decomposing the rational function using partial fractions, each term can be integrated individually using standard integration formulas. Typically, the integration of terms like 1/(x - a) leads to logarithmic functions, and terms involving quadratic forms such as (Bx+C)/(x^2+1) often result in expressions that include arctan functions. This approach leverages known antiderivatives for efficient evaluation of the original integral.

Transcript

00:02 Okay, so for us, we're trying to solve this integral, utilizing the form they give us down here.

00:09 So first, what we have to do is we've got to take a times x squared plus 1, and bx plus c times x minus 5.

00:31 And then we'll go from there.

00:32 So obviously up here, we have a x squared plus a down below, a little more, not really more, complicated, but we'll foil these together.

00:45 So i have b x squared minus 5, bx plus cx minus 5c.

01:01 Okay, now i want to set them up.

01:04 So we'll group squared, squared, we'll group x's with x's, and then constants with constants.

01:24 That my constant terms have to add up to two.

01:31 I know my x squared terms must add up to zero, and my x terms have to add up to 10.

01:39 So what i have, or what i end up having, is a plus b.

01:45 We'll kind of factor out the variables.

01:48 So a plus b, x squared, right? that equals 0 if i divide by x squared it's just as good as saying that it's not there right because i divide by x squared i end up still with zero on that left side so i'm just showing you how i got that those numbers so same thing with the negative 5b and the c will group those together negative 5b plus c those are my x terms and those have to equal 10 because there's a 10 as a coefficient out in front of that x and then finally finally, i have a minus 5c equals 2.

02:33 Okay, the easiest way i can think of to go about this is take our equations and kind of simplify.

02:40 So a really equals negative b...

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