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# Evaluate the indefinite integral. $\displaystyle \int \frac{dx}{5 - 3x}$

## $$-\frac{1}{3} \ln |5-3 x|+C$$

Integrals

Integration

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and this problem. We are practicing an integration technique called the U substitution. And this is a really helpful technique to solve. Integral is because what we do is we technically substitute the hard part of our integral to make our integral into something that we recognize that we can solve easily. And that's how we use U substitution. So that's what we're going to be doing in this problem. So let's just review the integral that we're starting with. We have the indefinite and the girl meeting. We don't have limits to plug in. We have the indefinite integral DX over five minus three x. Well, if we weren't using use substitution. Or maybe you solve the integral for the first time, that looks pretty hard. Maybe we don't know how to solve it. So we're going to use U substitution. We're going to let u equal five minus three x So we would take the derivative d u equals negative three d x so we can rearrange that and thio finding something that we know we have in our integral, which is d x so d x equal negative do you over three So we can rearrange are integral using the variable. You we would get negative 1/3 times the integral one over you and that looks like something that we can anti derive. That's a function that we know that we can solve. So we would get negative 1/3 times the natural log of the absolute value of U plus C remember sees that constant that we have to add on to an indefinite integral. So then we have to plug in our substitution for you. We can't just make this substitution and call it a day. We have to plug in that substitution back into our essentially our anti derivative to in order to get the correct answer. So we started with the integral DX over five minus three x. So once we plug you back in, you would get that Our solution is negative 1/3 times the natural log of the absolute value of five minus three x plus c. So I hope with this problem helped you understand how we can use u substitution to solve an indefinite integral on. I hope that it makes sense why we use U substitution in the first place.

University of Denver

Integrals

Integration

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