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Make a substitution to express the integrand as a rational function and then evaluate the integral.

$ \displaystyle \int \frac{dx}{x \sqrt{x - 1}} $

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$$2 \tan ^{-1} \sqrt{x-1}+C$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

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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.

06:41

Make a substitution to exp…

01:06

03:53

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06:56

01:24

02:12

04:38

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03:50

let's make a substitution such that after the substitution, this will be a rational function That means polynomial divided by polynomial. And then we'LL see if we can do maybe a partial fraction to composition. Maybe, Maybe not. Depends on the fraction. Here, let's try and take you to be X minus one in the radical. Yeah, or if you like, writers may be in one half, then, do you? We have one over to group X minus one D X. Now we can also write this, maybe push that to to the other side. And we have DX over radical X minus one so we can rewrite this so dx over the radical. That's just too, do you? But we have this extra ex term down here, but I can't write X here because, for example, I don't want to do that because I'm using a new variable You. So we have to lose the ex, come back to our substitution and solving for X so square both sides and that one over. And instead of writing X, we right, you squared plus one. So maybe this might be more recognisable if he plotted to. You may memorize this one. By now, if you haven't memorized it in the table, So let's maybe mention two ways to go about this. First, let's go ahead and just write that this is too ten and versatile. And the reason for this is because if you take the derivative attention inverse tangent, you may remember that from differential calculus. Or you could go ahead and actually just do it. Trips up here, you can let you be one tan data, and when you simplify, you should get ten in verse. Finally, let's get this answer back in terms of X so we just replace you by using the use of and that will be our final answer. So this equals to our ten radical X minus one, plus our constancy of integration, and that's our answer.

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