Evaluate the following integrals. ∫0^2 (x)/(x^2+4 x+8) d x |...

Name: Problem 32, Chapter 8: Calculus: Early Transcendentals
Uploaded: 2020-07-10T21:09:32-08:00
Duration: 5 min 14 s
Channel: Patrick Vaughn
Description: Evaluate the following integrals. ∫0^2 (x)/(x^2+4 x+8) d x

Key Concepts

Rational Function Integration

This concept involves techniques used to integrate functions that are ratios of polynomials. In many cases, especially when the degree of the numerator is lower than that of the denominator, the integral can be tackled using methods such as substitution or partial fractions after simplifying the expression appropriately.

Substitution Method

The substitution method is a standard technique in integration where a new variable is introduced to simplify the integrand. In integrals involving rational functions, a careful choice of substitution can transform the integrand into a form that is easier to integrate, often reducing it to a standard integral form.

Completing the Square

Completing the square is a technique used to rewrite quadratic expressions in a form that reveals certain inherent symmetries or standard forms. This method is particularly useful in integrals where the quadratic expression appears in the denominator, as the rewritten form often facilitates the application of substitutions or integration formulas such as those leading to logarithmic or inverse trigonometric functions.

Transcript

00:01 In this problem, we want to evaluate the integral of x over x squared plus 4x plus 8 on the limits from 0 to 2.

00:09 So we're going to use a completing the square on the denominator to get this in a form that's more workable.

00:17 So if we complete the square of x squared plus 4x, 4 over 2 is 2, so we'll want to use x plus 2 as the base.

00:27 0 to 2 x plus 2 squared now 8 if we take 2 squared away from that we'll get 4 and so 4 and x on the top stays the same dx next we want to use a u substitution on this where u is equal to x plus 2 that way we'll get u squared plus 4 on the bottom and we can use that later to a value a tangent inverse function.

01:02 So, u is x plus 2.

01:04 D .u is dx.

01:10 And our limits, we'll just add to two to each.

01:14 2 to 4.

01:16 If x plus 2 is u, then x is u minus 2.

01:22 Denominator, we have u squared plus 2 squared, just to make it easier to read this when we go to use our formula.

01:31 And that'll be in du.

01:32 So next, we want to split up the numerator into two integrals.

01:44 So first integral, 2 to 4 of u over u squared plus 2 squared du.

01:53 Second integral will bring the negative 2 out front, 2 to 4 of 1 over u squared plus 2 squared to you.

02:05 How the second integral is going to evaluate to our inverse tangent, but the first one we have to play with a little more.

02:15 We can use another substitution, v is equal to u squared plus two squared.

02:22 So dv would be 2 u, du.

02:30 Now we can rewrite these, so our limits are going to be 2 squared plus 2 squared.

02:37 That's 4 plus 4, so we get 8.

02:44 And up top, we're going to get 16 plus 4, so that's 20...

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