1-80 Evaluate the integral. ∫(x lnx)/(√(x^2-1)) d x |...

Name: Problem 58, Chapter 7: Calculus Early Transcendentals
Uploaded: 2021-12-11T20:46:12-08:00
Duration: 8 min 42 s
Channel: M Z
Description: 1-80 Evaluate the integral. ∫(x lnx)/(√(x^2-1)) d x

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique in integral calculus used to integrate the product of two functions. The method is based on the product rule for differentiation and is particularly useful when one of the factors, such as a logarithmic function, differentiates to a simpler form while the other factor can be integrated with relative ease.

Trigonometric Substitution

Trigonometric substitution is a method for evaluating integrals involving square roots of quadratic expressions, such as ?(x² - a²). By substituting x with a trigonometric function (for example, x = sec(?)), the radical is transformed into a trigonometric expression that is often easier to integrate. This technique is essential in handling integrals with forms that are otherwise resistant to standard algebraic manipulation.

Handling Logarithmic Functions in Integration

Logarithmic functions, like ln(x), require special strategies when integrated within a product. Their differentiation typically simplifies the integrand, which makes them well-suited for techniques like integration by parts. Understanding how logarithmic functions behave under differentiation and integration is critical for tackling integrals where these functions are present.

Transcript

00:01 Okay, so this is the integral that we'd like to look at.

00:05 And we're going to do this using integration by parts.

00:10 So what are the parts that we're going to pick? well, first we'll let you equal to lnx.

00:17 And then our dv is going to be the rest.

00:20 So our dv is going to be x over square root of x squared minus 1, dx.

00:29 Okay, so then, well, we can easily say that du is equal to 1 over x, dx, and then just taking the derivative of both sides.

00:38 But to find this v here, well, we need the integral of this x over square root of x squared minus 1, and that's a little bit tricky.

00:48 So let's come over to the side and do that separately.

00:53 So integral of x over squared of x squared minus 1, oops, x squared minus 1, dx.

01:05 Well, let's do this using u substitution.

01:09 So let's let you equal x squared minus 1, and then du is going to be equal to 2x dx.

01:18 And then let's bring this two over the other side, so we've got that one half du is equal to x.

01:23 Dx.

01:26 And then continuing on using this u sub, well, we can replace this x, dx with one -half du.

01:37 So let's bring our one -half outside.

01:41 This is, and we can substitute in, so this is going to be the one -half integral of square root of u, d -u, because this whole x -d -x was replaced with this one -half d -u.

01:56 And actually, i'm going to revise what i said, and i'm actually going to keep this one -half on the insides.

02:03 I'm going to write this as one -half times one over square -a -u -d -u.

02:08 Why i'm going to do that is because, well, 1 over 2 square root of u, that's just the derivative of square root of u.

02:14 So we can say that this is just equal to square root of u plus c.

02:20 And then now we can just substitute back in.

02:22 So this is just equal to squared of x squared minus 1 plus c.

02:29 Great.

02:30 So what have we found? okay, well, all this was just defined what our v was equal to.

02:33 So we just want the answer to an antide derivative of this.

02:36 The v is just equal to x squared minus 1.

02:42 Right.

02:42 Okay, so now using this, well, now we can apply the integration by parts.

02:48 This will be equal to our u times v.

02:51 So that's ln of x times squared of x squared minus one minus the integral of our vdu.

03:02 So that should be square of x squared minus one over x d x.

03:10 Okay, but now what do we do with this guy? so this is also pretty, not a super obvious integral.

03:17 So let's come over to the side and look at doing this integral.

03:24 So let's see, let's shift all the way over here, because this is going to be a bit lengthy.

03:31 This is the integral that we want to do here.

03:33 So the integral of square root of x squared minus 1 over x, dx, so how are we going to do this? well, again, this one is kind of tricky.

03:44 So let u equal to square root of x squared minus 1.

03:51 And then du is going to be equal to 1 over 2 square root of x squared minus 1.

03:59 And we need to do the chain rule.

04:00 So we need to multiply this by 2x and then dx.

04:04 All right.

04:05 So then, well, we can rewrite this elastic quality that we just wrote.

04:10 In a way that's a little bit easier.

04:12 So a little bit nicer.

04:14 Well, first off, we can just cross off these two u's...

Please give Ace some feedback