Evaluate the indicated integral. (a) ∫(3 √(x))/(1+x^3) d x (b) ∫(x √(x))/(1+x^5) d x

Name: Problem 26, Chapter 4: Calculus: Early Transcendental Functions
Uploaded: 2022-11-15T01:12:45-08:00
Duration: 2 min 54 s
Channel: Elliott Walker
Description: Evaluate the indicated integral. (a) ∫(3 √(x))/(1+x^3) d x (b) ∫(x √(x))/(1+x^5) d x

Evaluate the indicated integral. (a) ∫(3 √(x))/(1+x^3) d x...

Key Concepts

Substitution Method

Substitution is an integration technique used to simplify integrals by changing variables. In many cases, especially when the integrand is a composite function where the differential of an inner function is present, substituting that inner function with a new variable can transform the integral into an easier or standard form.

Inverse Trigonometric Integration

Inverse trigonometric integration involves recognizing integrals that take the form of standard inverse trigonometric functions. A classic example is the integral of 1/(1+u^2), whose antiderivative is the arctan function. This technique is key when the substitution in an integral reduces it to such a standard form.

Transcript

00:01 Here we have two different indefinite integrals that we'd like to evaluate using integration by substitution.

00:04 So for this first integral, we have the integral of 3 squared of x divided by 1 plus x cubed d x.

00:10 To evaluate this, we'll go ahead and make the substitution u equals x to the 3 halves.

00:15 This means that x is going to be equal to u to the 2 thirds, therefore d x is going to be 2 thirds u to the negative 1 third.

00:23 So you can rewrite this integral, it's a definite integral of 3.

00:27 Square root of x or squared of u to the two thirds is going to be u to the one third and divided by one plus x cubed or one plus u squared then d x we can rewrite that as two thirds u to the negative one third d u so there's a d u term there as well this is equivalent to the integral our three and three cancel out uh leaving us with a factor of two which we'll put outside u to the one third times u to the negative one third is just one so those cancel each other as well so it's two times the definite or the indefinite integral of 1 over 1 plus u squared d u now this has an important form this is the form of the derivative of the arc tangent so we can say this is equivalent to 2 times the arc tangent taking the anti -derivative of u plus c don't forget your arbitrary constant we want an derivative in terms of x so we'll go ahead and just substitute what we know u is equal to x to the 3 halves so this is 2 arc tangent of x to the 3 half so this is 2 arc tangent of x to the 3 halves plus c.

01:30 It's the value of this.

01:31 You can take the derivative of this to confirm that is indeed equivalent to this original integrant.

01:37 Now for b we'll make a similar substitution.

01:38 This time we'll use u equals x to the five halves because we have the antiderivative of x times the square of x over 1 plus x to the fifth d x...

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