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Evaluate the integral.

$ \displaystyle \int_0^1 \frac{3x^2 + 1}{x^3 + x^2 + x + 1}\ dx $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

Missouri State University

Baylor University

University of Nottingham

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

04:34

Evaluate the integral.…

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04:29

00:42

Evaluate the indefinite in…

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01:20

for this integral. Let's use partial fractions. So do that. We really should look at the denominator to see if this factors so this what you could do here they see that X equals minus one is a route. So if you plug it into that denominator, you get zero. So this tells you the X minus negative one or just X plus one divides the denominator. So go ahead and do the long division there. So you would like to divide over X Plus one. And after doing the polynomial division, we X Square plus one. So that tells us that we can go ahead and write the original and so grand and the denominator. We can now write that as X plus one times x squared plus one. And we get that by just multiplying this to both sides. And then we'LL get a factor on the right so we can replace this cubic polynomial with X square plus one Times X plus one. So now we would do the partial fractions. This's a over X plus one bye B X plus C explored plus one. Now go ahead and multiply both sides of this equation by the denominator on the left. Then we get a and then b X plus C with X plus one. Now go ahead and simplify this friendly hand side. And we could rewrite this by pulling out of X Square by pulling out of X and then we're left over with our constant term. So look at the coefficients on the left we see a three in front of the X Square, so a plus b is three. Then we see no ex term over here on the left. That means D plus he must be zero. And in the constant term is one. So the constants of here must be want that gives us three by three system to solve. Come. So here, for example, you could be equals negative c and then we can write This's a minus. B equals one and then plug this equation and right below we can go ahead and add these equations together. You get ankles, too. From this equation up here, B equals one. And then from here we have C equals negative one. So let's go ahead and on to the next page. And let's plug in these values for a B and c into our partial fraction. So it was to be was one. So we just have a one x there and then see was minus one. So we get the mine. Is there exploring? Plus one. Now let's split this into three in a girls Then for the second one, we have X and then finally So for this first integral if this plus one is bothering you, do you sub? And then when we integrate that we'LL get to natural log absolute value It's close one and point zero one for the second Integral Here you Khun do another use, um let you be the denominator Then do you over two equals x t x So there we'LL get one half natural log explorer plus one zero one and then finally over here You may remember this one already, but if not, you can do it Tricks up here X equals tan data after using that you should get But now including this minus here will have minus Artie Innovex zero one and then we just plug in. So here's the first one plug in one for X to national log to when you plug it zero for X you get natural other one, which is zero. And then over here, plug in X equals one. Plug it zero than the whole term zero and then ten inverse of one. And then plus, because of this minus here, it's an inverse of zero. So this is just five over to natural log to combining these and then this is minus pira for And this is because this is over. Here is power for in this term over here. Zero. So we have the minus right there, and that's your final answer.

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