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Find the general indefinite integral.

$ \displaystyle \int (5 + \frac{2}{3}x^2 + \frac{3}{4}x^3) \, dx $

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00:41

Frank Lin

00:29

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Campbell University

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

00:47

Find the general indefinit…

02:19

Find the indefinite integr…

01:05

Alright We're tasked with the integral of five. Yes sure. Two thirds execute. It's squared. I said that wrong. Plus three force X cubed dx. And the theme in this problem is that when you're asked to do the integral of X to the end power, I'm just going to give you one rule because it's the same rule over and over again. You add one to the experiment and I like to multiply by the reciprocal of the new excrement. And what we learn is that you can do each piece individually so there's no expert on this this first one. So you just add an excellent uh I guess you can think of as adding X to the first and then divided by one. Um But it's the same thing that you're a diva five X. Is five. So what I like to do is just add one to the excrement and then multiply by the reciprocal and just think about if you took this value multiplied by 1 3rd. that three is going to go to the denominator to give you two nights. And you do multiply same thing with this next one is you add one to your experiment. But they have to multiply by your new exponent in four times forward. Give me 16, you have three and then don't forget about plus C. I guess. You forgot to write plus C over here because the derivative of a constant is zero. So we need to write that down and you can double check this is correct because it's derivative is the same thing as what's up here. So we're right.

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