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(a) Compute $\frac{d}{d x}\left(\frac{1}{b} \ln (a+b x)\right) ;$ use this to determine $\int \frac{1}{a+b x} d x.$

$$\frac{1}{a+b x}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

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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.

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I like the way this problem set up because they are basically giving you the answer and looking at this eso when they ask you to do the derivative of one Overbey natural log of a plus B X and what they're asking you to do is use the chain rule. Um, and what I'm talking about is that you do the first of all you leave the be an a r uh, constants. You leave one over, be alone. I thought it change colors. One over b. It is left alone times. The derivative of natural log is one over that quantity a plus b X. But then the changeable says that you have to take the derivative of a plus B x on you multiply by that derivative well, the derivative of a zero. So you don't have to write that down. But the derivative a B X is B, and what you'll notice here is that these pieces canceled. Eso alright, it's equal to this. So the whole premise of the problem that is, is if I asked you for the integral of one over a plus B x the X, then what we need to do is go backwards because the integral is the anti derivative. So if we go backwards and the answer to this, uh, well, because if you if you see we have one over a plus B X So the answer is one over be because I'm just rewriting the derivative, um, natural log of a plus B X. Or I guess I should say I'm rewriting the thing I took the derivative of, um but don't forget about a plus c in the problem. Um, the reason why you need a plus C is because the director of a constant is zero. So you might have, like, a plus zero over here. Eso This should be your answer.

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