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Problem

Evaluate the integral. $ \displaystyle \int \f…

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Problem 21 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int \arctan \sqrt{x}\ dx $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Related Topics

Integration Techniques

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01:53

Integration Techniques - Intro

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.

Video Thumbnail

27:53

Basic Techniques

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

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Video Transcript

Let's try a U sub for this integral. Let's take you to be square root of X so that do you one over to X so we can rewrite this as t U equals one over to you dx and then just saw that for DX. So this is what will replace this term over here dx within the next in the raw. So this integral equals ark tan of you and then here for D s. We have to you, do you from this equation over here. So now is pull into two war and then let's use integration by parts. So here, let's use a different letter than normal. Let's use w let that be ten inverse of you then DW one over one plus you square, do you? Then we have a TV must be was left over, which is you. Do you then integrate that to get use were over too. So here we're using the formula. So we have to and then recall the buy parts formula in this case because I'm using the letter w So here, let's go ahead and multiply out this to to both herbs. So we'LL have you swear ten in verse. You, after cancelling out two twos minus integral use, were over one plus you squared and then do you? So now we have another integral to deal with here. So perhaps the easiest way to do this one is to do polynomial division. And then you can rewrite this. After doing the long polynomial division, you can go ahead and write. This of the quotient is one and the remainder is minus one, so we can rewrite it. Is this so let's go to the next page. And then we have one minus one over one. Plus he square and this is a much easy, easier in a role to do. There's no need to do partial fractions here. This already is a partial fraction. So for this integral, you may remember this one. Well, we actually use this fact already the derivative of ten in verses, one over one plus u squared or if you completely for about this fact, you can just do a trick sub. In either case, this hole is fresh in here and then we have minus you. And then, because of this double minus plus ten in verse, you and then go ahead and add that constancy and recall that we defined you to be scared of X. So let's go ahead and chlo that in here. So we have used square, which is X here. Technically, you should put absolute value. But in order for this to even make sense, this definition appear we need X to be bigger than or equal to zero. So we could just replace you square with X. So no absolute value necessary there. And then here we have square X, Linus, Rolex Plus ten in an ember through X and then plus e. And then here you could go ahead and factor out Ah, and first hand if you like, and that's a final answer.

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University of Nottingham

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Calculus 2 / BC Courses

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Video Thumbnail

01:53

Integration Techniques - Intro

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.

Video Thumbnail

27:53

Basic Techniques

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

Join Course
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