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JH
Numerade Educator

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Problem 14 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{1}{(x + a)(x + b)}\ dx $

Answer

$$
\frac{1}{b-a} \ln |x+a|+\frac{1}{a-b} \ln |x+b|+C, \quad a \neq b
$$

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Video Transcript

let's evaluate the integral of one over X plus any times X plus B because Andy or could be any real numbers, this problem will depend on whether or not and your equal, because if a equals B, for example, this means that we can write the general as one over X plus a square. And that will change how we go about the partial fraction to composition. This will be something that the book will call case too. But really, for us in this case, this basically is already in the form of a partial fraction to composition. We just could actually go ahead and integrate this with the use of used the power rule. So this is why the case, it matters whether or not a equals B and we'LL see in case, too, when a does not equal b we'LL have two distinct linear factors in the denominator. So that's a different looking partial fraction the composition what we have over here now we're just circling right now. So first for in this case, a equals B we rewrite integral and then you may be able to integrate it how it is and once that. But if this plus a is bothering you just go ahead and do it yourself. Then you get do you equals the X. Okay, so we could write this integral is one over you square, do you? And then you could use the power rule to get negative one over you. Plus he and then go ahead, replace you with X plus a. So, in case one, if a equals B, this should be the general. Now we have to also look at the case in which there aren't equals. Let's do this next. So that's case one all over here. Boxed off in green. Now for case, too. And you're different. Distinct, not people. So now we can go ahead and take the original Fraction one over. That's plus a this this form right here as its in with A and B distinct. This is not a partial freshen. The composition here will be in with the book calls Situation one or case one it went. The denominator has this thing's in factors that are linear. So we'll have capital a over explosive plus capital B over X plus little bee. Now, looking at this latest equation, let's circle is in black. Let's go ahead and multiply both sides by the denominator on the left X plus little eh X plus little bee. Good and multiply both sides by that. And we should get one on the left capital. A Times X plus little bee plus capital B Times X plus little and the right. Let's fact, throughout the X on the right hand side, we have a plus B, and then the constant term is capital a time. Small B plus capital B time. Small eh? Now, on the left hand side, we see that there's no ex term. This means that the ex term, the coefficient of the extreme on the right must be zero. And that's equivalent to saying B equals minus a. And then, since the constant her among the left is one. That means that the constants room on the right also has to be one. And now, using the fact that B equals minus A, we could come back to this equation up here and saw for large, eh? So it's fact throughout a large a Here and divide we get one over little B minus little, eh? And then, since being equals minus a due to this equation Over here we have negative one over B minus some running on room Here, let me go on to the next page. So the next thing that will do it before we go on is we'LL take that. We're ready to integrate We'Ll take this original problem here. We wrote it in partial fraction of composition and now we just found a and B So it's about unplugging these values for Andy into the perfect reaction to composition. Yeah. So this means that the original problem Let's write that out. And then we found large, eh? Which is one over B minus a. That's the constant capitally over explosive A And then we also found these Hey, B was negative one over B minus A. Oliver explodes be. And this these terms are easier to generate on the first integral and might help you to do it. Use up U equals explosive. For the second one, you can go ahead and take new equals X plus Meet Teo. Integrate these And when we integrate, we should have won over B minus a natural log. Absolute value of you are explicit and then we have minus one over B minus A for the next integral. That's minus is coming from this term up here and then natural log. Absolute Value X plus B. That's our integration. And then, since this is a definite over all week putting to sea, the last thing to do here is just to simplify. We see that we can pull out a one over B minus a. From the first two terms. You have Ellen X Plus a minus Ellen X plus me, those air an absolute values and then the last thing we can do. Go ahead and use your log properties to rewrite this determined Prentice's. We have won over B minus a natural log, absolute value explosive over X plus me plus see, and there's a final answer.