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Evaluate the given integral and check your answer.$$\int\left(t^{2}+1\right)^{2} d t$$

$$\frac{t^{5}}{5}+\frac{2 t^{3}}{3}+t+c$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

Missouri State University

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University of Nottingham

Boston College

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.

01:17

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02:03

all right. We're looking at the integral of this function T squared, plus one squared. Now, simplifying this problem at the beginning is what you want to do. So just a reminder off to the side. I'm just gonna remind you that squaring something means multiplied by itself. So as you distribute that t squared in and becomes T to the fourth plus one, times T squared is t squared and then distribute this one in. So we have another t squared and one times wanted to give me one so I would make my students rewrite this as in simplified form T to the fourth, we have two t's words plus one. I combine like terms here in the middle. DT. So now I can go ahead and do the anti derivative which is adding one to the exponents and multiply by the reciprocal that new exponents add one to the exponents on this time, I'm just going to divide because multiplying by the reciprocal the same thing is dividing by that new explanation plus t because that's an anti derivative of one. And then don't forget about a constant. And when you check your work, you're taking the derivative of this. So just with a different color. If you take the derivative of this, bring the five in front that will cancel. Subtract one from the exponents. You get T to the fourth. Bring the three in front in the 30 Cancel your left with two. Subtract one from your exponents. T squared. A derivative of tea is one the derivative of a concept zero. There's no need direct plus zero. Yeah, so that's proving checking our work that my answer and green is correct.

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