Evaluate the integral. ∫(2 x^5-x^3-1)/(x^3-4 x) d x |...

Name: Problem 20, Chapter 9: Calculus A New Horizon
Uploaded: 2021-07-13T16:58:12-08:00
Duration: 9 min 27 s
Channel: Benjamin Kelly
Description: Evaluate the integral. ∫(2 x^5-x^3-1)/(x^3-4 x) d x

Key Concepts

Polynomial Long Division

This is a method used to simplify rational functions when the degree of the numerator is equal to or higher than the degree of the denominator. By dividing the numerator by the denominator, the function is expressed as a polynomial plus a proper fraction, which often simplifies the process of integration.

Partial Fraction Decomposition

This technique involves breaking down a complex rational expression into a sum of simpler fractions whose denominators are factors of the original denominator. In integration, these simpler fractions are usually easier to integrate, often reducing the integral to a standard form.

Factorization

Factorization is the process of expressing a polynomial as a product of its factors. In the context of integration, factoring the denominator (and sometimes the numerator) is crucial for setting up a partial fraction decomposition, thereby enabling easier integration of each term.

Standard Integration Techniques

Once a rational function is decomposed into simpler fractions, standard integration formulas can be applied. This includes basic integrals such as the natural logarithm for 1/x-type functions and power functions, which provide the antiderivatives needed to solve the integral.

Transcript

00:01 So in this problem, we're asked to evaluate this integral, and we see we have a higher power term on the top in the numerator than in the denominator, because 5 is greater than 3.

00:11 So why don't we start with some polynomial long division? so we'll divide 2x to the 5th minus x cubed minus 1 by x cubed minus 4x.

00:22 So the first thing we want is to get 2x to the 5th plus something else.

00:28 So we'll multiply x cubed minus 4x times 2x squared.

00:33 So 2x squared times this is equal to 2x to the 5th minus 8x cubed.

00:42 So let's subtract this and we get 0 plus 7x.

00:49 And we can drop this one down minus 1.

00:53 And so now we want to get 7x cubed.

00:55 Through multiplication.

00:57 So if we multiply x cubed times seven, we get 7x cubed.

01:01 So put that there and multiply.

01:05 We get 7x cubed minus 28x.

01:10 And so now we can subtract.

01:12 We get 0 plus 28x minus 1.

01:16 And so now this is of a higher power than of this.

01:21 So we'll stop there.

01:24 So what this means is that we we have 2x to the 5th minus x cubed minus 1 is equal to x cubed minus 4x times 2x squared plus 7 plus, and this is our remainder, 28x minus 1.

01:45 And so going back to our original integral, we had 2x to the 5th minus x cubed minus 1 over x cubed minus 4x.

01:55 And so this is equal to this all over x cubed minus 4x and this over x cubed minus 4x okay so now it'll be a little easier we can do partial fraction decomposition on this term and so we have 28x minus 1 over x cubed minus 4x which is equal to, we can simplify the bottom to take out an x, we get x squared minus 4.

02:33 And so this is a perfect square, so we can simplify even further.

02:39 It's equal to x times x plus 2 times x minus 2.

02:44 So now we can do the partial fraction decomposition.

02:48 So we have 28x minus 1 is equal to a over 28.

03:02 X minus 1 over x cubed minus 4x is equal to a over x plus b over x plus 2 plus c over x minus 2 and so now we'll multiply each term by x cubed minus 4x or by this here and we get 28 x minus 1 is equal to a times x plus 2 x minus 2 plus b times x times x minus 2 plus c times x plus 2 okay well now we can multiply these out you get 28 x minus 1 is equal to a times x squared minus 4 plus b times x squared minus 2x plus c times x squared plus 2x and so now why don't we distribute a, b, and c into these parentheses, and we get 28x minus 1 equals ax squared minus 4a plus bx squared minus 2bx plus cx squared plus 2cx.

04:34 X's and just a.

04:37 And so we can sort these terms to where we have 28x minus 1 equals x squared times a plus b plus c plus x or actually 2x times c minus 4a.

04:59 And so now we have 28x minus 1 to all of this.

05:07 So since we just have to solve for coefficients now.

05:12 So since we don't have an x squared on the left side, we can say that a plus b plus c is equal to zero.

05:19 So we have 28x...

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