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Evaluate the integral.

$ \displaystyle \int \frac{5x + 1}{(2x + 1)(x - 1)}\ dx $

$$\frac{1}{2} \ln |2 x+1|+2 \ln |x-1|+C$$

Integration Techniques

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Oregon State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Let's evaluate the following integral and let's go ahead and take the fraction inside the inte grant and go ahead and rewrite this using partial fraction decomposition. This is what the textbook would call case one. We have distinct linear factors in the denominator, so we can rewrite. This is a over two X plus one plus B over X minus one and then let's go ahead and multiply both sides by the denominator two X plus one x minus one. So we multiply on the left, the two X plus water and the X minus ones will cancel. So we're left with five x plus one on the right, where we multiply by this fraction by this product of linear polynomial. The two x plus ones will cancel when we multiply A. So we have a X minus one and then similarly for B, the X minus ones will cancel. And so we left over with two X plus one. So let's go ahead and rewrite the right hand side. So let's factor out an X here, and then we have a and then to be, and then we have B minus a. So if these two expressions are equal, then that must mean that the terms in front of the X term are equal and also that the constant terms are equal. So this gives us two equations. We have a plus to be equals five. But then we also have that B minus A is one. So we have a two by two system for A and B. Let's go ahead and solve that many ways To solve this one way is to just take this equation over here in green, solid for B and then go ahead and plug this be value into the other equation. So if we do that a plus to be replace, be with one plus a this is equal to five. But then we can go ahead and simplify this. We have three a equals five minus two. So that means a is one. And then using this equation over here, we also get that B as to so I'm running out of room here, let me go into the next page. So now the next thing, actually, before that what we should do here is go ahead and replace a and B with the values that we found. So now We'll replace A and B with one and two and then we'll integrate. Instead of integrating the left hand side, we'll go ahead and integrate the right hand side. So we have a was equal to one and then be was equal to two and now we have two. Integral to deal with. It may help you here to use a U sub. If you'd like to use the use up here, you can let you be two x plus one for this integral here. If you don't like that minus one on the X, you can go ahead and do another sub here. Let's do w this time then D w equals d X do u equals two d X so you can go ahead and carry out the U substitution here to evaluate this Integral. In either case, we are going to see natural algorithm. So for the first in overall, we get the one half natural log Absolute value two x plus one and for the second, integral to natural log Absolute value, X minus one plus our constancy. And that's our answer