Evaluate the following integrals. ∫(x^3+6 x^2+12 x+6)/((x^2+6 x+10)^2) d x

Name: Problem 59, Chapter 8: Calculus: Early Transcendentals
Uploaded: 2020-05-28T16:08:37-08:00
Duration: 11 min 45 s
Channel: Tom Greenwood
Description: Evaluate the following integrals. ∫(x^3+6 x^2+12 x+6)/((x^2+6 x+10)^2) d x

Evaluate the following integrals. ∫(x^3+6 x^2+12...

Key Concepts

Integration of Rational Functions

This concept involves techniques for evaluating the integral of a ratio of polynomials. It is an important topic in calculus that often requires methods such as polynomial division, substitution, and partial fractions in order to rewrite the integrand into a form that can be integrated using standard formulas.

Partial Fraction Decomposition

Partial fractions is a method used to express a rational function as a sum of simpler fractions. This technique is particularly useful when the denominator can be factored or written as repeated irreducible quadratic factors, allowing for the individual integration of each term.

Substitution Method

Also known as u-substitution, this technique simplifies the integration process by changing variables, which can turn a complicated integrand into a simpler one. When part of the integrand has its derivative present elsewhere, substitution can help to reduce the integral to a standard form.

Completing the Square

This algebraic technique is used to transform a quadratic expression into a perfect square plus or minus a constant. It is often applied when dealing with quadratic expressions in the integrand, especially in the denominator, to facilitate substitutions or to match standard integral forms.

Polynomial Division

When integrating rational functions where the degree of the numerator is equal to or greater than the degree of the denominator, polynomial division is used to divide the numerator by the denominator. This process separates the integral into a polynomial part and a proper rational function, which can then be integrated separately using other methods.

Transcript

00:03 Okay, we're trying to do this integration, and the first thing we have to do is split this fraction up using the partial fraction decomposition.

00:11 I'm not going to rewrite the whole fraction here.

00:14 So just because we have a repeated quadratic form, there's going to be one fraction with an x squared plus 6x plus 10, just the first power in the denominator.

00:27 And another fraction that will be the x squared plus 6x plus 10 to the second power.

00:37 And each of these has to have a numerator with a x term and a constant term in it.

00:42 So that will be a x plus b on the first one, and then cx plus d on the second one.

00:51 Then to get our decomposition system of equations here, first we have to multiply everything, by the common denominator.

01:01 So that will give us just our numerator in the original fraction, the x -cube plus 6x squared, plus 12x, plus 6.

01:11 Now on the other side, when we multiply each of these fractions, we're going to be able to do some reducing.

01:19 And the first fraction, the ax plus b, only one of the quadratic terms is going to factor out.

01:26 So we have a x plus b times, the x squared plus 6x plus 10.

01:33 And so when we multiply that out, just doing the a x times the whole thing and then the b times the whole thing, we get a x cubed plus 6 a x squared plus 6 a x squared plus 10 a x.

01:55 And then taking the b times each of that, we get plus bx squared plus bx squared plus 6b times x plus 10b and then when we take the second fraction the entire common denominator will divide out and we're just left with the plus cx plus d let me get mine like terms lined up there so now we can equate the terms on each side and write our system of equations so we have our x cubed has to equal our x cubed.

02:39 So, a is 1.

02:42 And then we have our squares have to equal each other.

02:47 So our 6x squared would have to equal the 6a plus b, x squared.

02:55 And since a is 1, that means that b equals 0.

03:01 And then our x term, the 12x, they have to equal all of our x terms over here added together.

03:09 So that's going to equal 10a.

03:13 Since b is zero, i'm not going to bother writing that plus c.

03:18 And that tells me that c has to equal two because a is one.

03:24 And then finally, our constant term, the six, has to equal our constant terms of, again, b is zero.

03:34 So six equals just d.

03:38 So there are the coefficients that i'm going to have in my partner.

03:42 Fraction decomposition.

03:45 So that means this integral can get rewritten as the integral of a x plus b, so just 1x over the x squared plus 6x plus 10 plus the cx plus d.

04:07 So that'll be 2x plus 6x over the x squared plus 6x plus 10 squared, and that all is multiplied times the dx.

04:24 Now, this first fraction is really the one that causes us all the problems.

04:29 If you saw me do a trick earlier where i separated my constant, so i'd have a perfect square trinomial.

04:36 That's not going to work because that's going to give me x plus 3 squared.

04:43 And if that's my u, d .u doesn't have an x.

04:46 It and we have an x in the numerator and i can't just let x equal the entire denominator and try my or let u equal the entire denominator and just try and do my one over u d u because then that would have a constant term with it and my numerator doesn't have a constant term to go with the du part so again we have to do a little algebraic trickery and we have to rewrite this numerator as another expression that will have the x term and the constant term with it.

05:25 And so to do that, i'm going to multiply this x by a 2, and i'm going to add a 6 to it, because that will make it look like the derivative of the x squared plus 6x plus 10.

05:45 That will give me 2x plus 6...

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