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Evaluate the integral.

$ \displaystyle \int \frac{2x - 3}{x^3 + 3x}\ dx $

$\ln \left|x^{-1}\right|+\ln \left(\sqrt{x^{2}+3}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

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let's use a partial fraction the composition for this one. So first, before we do that, we should check if we can factor the denominator. And in this case, you can pull out of X left over with X squared plus three. Now, I would also have to check it this quadratic here in Prentice's if that matters, so you can try to factor this good. This one doesn't factor. And we get this by looking at the discriminative B squared minus four a. C. And this problem. There's no ex term in the quadratic. So be zero and then minus four times the coefficient in front of X squared is one. And then times three, which equals C. And this is a negative number. So that means that this quadratic will not factor. So using the case won for the linear factor. This's what the author calls case one non repeatedly near factor. And then here This is what the other calls case three. Because it's irreducible, quadratic. So in the numerator we need be explosive this time. And then that denominator X squared plus three. So now it's more supplying both sides of this new equation By that denominator on the left. So on the left side we have two ex ministry, but on the right, we have that. And then we can go ahead and simplify this. So let's pull out of X squared and we could pull a plus B and then we have CX and then we have three, eh? So I just combined these terms by factoring out some power of X. So here, x square, Then extend the constant term. And the reason for doing this is that we look at the left hand side. We see that there's no X squared over here, So zero x square. So on the right hand side and next to the X square, that's her must also be zero. So a plus B equals zero. Similarly, we have C equals two and then we have three A equals minus three. So this is two over here and then minus three over here. And so go ahead and saw these three. So here we have a CZ minus one. Then plug this into this and then you get B is one. And here we're from this equation we no see is too. So let's go ahead and plug this A, B and C, and so are partial. Fracture the composition. And then instead of integrating the left hand side, the original problem book integrate this right hand side. So let me go to the next page to write this. So a was negative one so minus one over X and then be was one. So one X and then see was too. So plus two, that's our inaugural. One way to proceed is to just right this into split this into three intervals. So I'm getting a little sloppy. Here, let me take a step back. Negative one over X. That's the second one. And then for the last one, we could even pull out that too. All right, so three in a girl's here, the first one that's just negative. Natural log, absolute value of X. For the second and roll. We can do a use of let you be that denominator. Then, after that, if you do to you and then divide by two, you get extra e x, which is the numerator. So this I have some use over here. So one half and then let me just write this in terms of X after you do the use of and right and back substitute. That's what you have. And then finally, for this last inaugural you saw was not going to work because of this extra X that we get the numerator. So for this one, we can do it tricks up X equals square root three tan data, then DX route three c can square data. So after doing this, tricks up for the third and rule and then rewriting everything back in terms of X, we have two, three over three and then we would have Seita. But then instead of so here, we would have a date after we integrate. But then we have ten data people's eggs over root tree. This is for martyring sub and then you could solve for data here by taking the are ten. So you would replace data and this expression here with our ten. And so let's write that and let me also finally let me add that seeing since we did integrate and then one half and actually on the second log, you could even drop absolute value since X squared plus three is always positive. And then to room three over three. And now we replace state of with this and boom, add that constancy and that's our final answer

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