1-80 Evaluate the integral. ∫(d x)/(x+x √(x))

Name: Problem 55, Chapter 7: Calculus Early Transcendentals
Uploaded: 2021-12-10T04:04:12-08:00
Duration: 4 min 30 s
Channel: M Z
Description: 1-80 Evaluate the integral. ∫(d x)/(x+x √(x))

Key Concepts

Algebraic Manipulation

This involves rewriting and simplifying expressions by factoring common terms or combining like-terms to make an integral more tractable. For example, factoring may reveal opportunities for substitution or other techniques that simplify the integrand, an essential skill in approaching complex integration problems.

Substitution Method

Substitution is a technique used to simplify integrals by changing variables, effectively transforming the integrand into a simpler form. It is particularly useful when the derivative of a part of the integrand is present elsewhere in the expression, allowing the integral to be rewritten in terms of a new variable.

Partial Fractions Decomposition

Partial fractions are used to break complex rational functions into a sum of simpler fractions, which makes the integration process more straightforward. By expressing a complicated fraction as a sum of simpler components, each term can often be integrated individually using standard methods.

Integration of Rational Functions

Integration of rational functions involves techniques for integrating fractions where the numerator and denominator are polynomials. This typically requires a mix of algebraic manipulation, substitution, and partial fractions to reduce the function to forms that are easily integrable with basic calculus formulas.

Transcript

00:01 Okay, so the integral that we would like to do is this integral of 1 over x plus x square root of x dx.

00:10 I'm going to start by rewriting this actually as 1 over square root of x times square root of x plus x.

00:19 And you can see if you distribute the square root of x, you're going to get the same thing back that you started with.

00:24 So i'm going to start with a u sub.

00:26 So let's let u equal square root of x.

00:30 Taking the derivative of both sides, we're going to get that du is equal to 1 over 2 square root of x dx.

00:37 And now this is going to actually be able to tell us that this is just rewriting this.

00:42 This is saying that 2du is equal to 1 over square root of x dx.

00:48 Okay, great.

00:48 So now continuing on, so we're going to substitute in.

00:51 So this is equal to the integral of, well, so this is 1 over x here, is actually going to be replaced now by 2du.

01:06 Because now we have this 1 over x, we're just going to be able to replace this 1 over x dx with this 2d.

01:12 Okay, so what do i mean by that? so now we're going to get this is equal to the integral of 2 over, and now we're going to replace this square root of x with a u.

01:24 That's the sub that we made.

01:25 And then this x here is, well, it's use.

01:28 Squared because if u is squared of x then u squared is just x and then just d u at the end okay and then i'm actually going to rewrite this a little bit as um two times the integral 1 over u times u plus 1 d u and then now how are we going to do this well the way that i think uh is best to approach this uh this integral here is using uh partial fractions so let's come over to the side here and figure out what the partial fraction is for that fraction.

02:11 Let's see.

02:14 Partial fraction.

02:15 Okay, so we want 1 over u times u plus 1...

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