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Evaluate the integral.

$ \displaystyle \int \frac{4x}{x^3 + x^2 + x + 1}\ dx $

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

Integration Techniques

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let's evaluate the given integral. So looking in the denominator, we see a third degree polynomial and we should factor the city factors. But it might not be obvious had a factory. So what you might try here is the rational zeros test, which basically says if you have a polynomial within surgical efficient CE, which is what we have here, you look at all the factors of the constant term in this case, just plus and minus one, you look at the leading terms coefficients also plus or minus one. And then you divide these terms by these serves and these air all your possible rational number solutions. In this case, if we plug in X equals minus one, it's like that into the part of your you get negative one cubed, which is minus one negative one squared, which is one. So those cancel plus X, which is negative one and then plus one. So those negatives cancel up and we get zero. So this tells us the X minus negative one or X plus one divides this polynomial so we can go ahead, perform the long division to see the other factor. So we have X plus one out here we do. Our long Division X three over X is just like squared scoring. Subtract where some cancellation here, X plus one x divided by X, is one and then we get X Plus one and we should get her remain your zero. That's what we get. So this tells us that we can, right, The denominator is X plus one and then X squared, plus one so much easier to deal with. And then once more, we go to this quadratic terms and we'd like to see whether or not that factors So here you just check the discriminatory B squared minus four a. C. We're here the A, B and C. These are just the coefficients of arbitrary polynomial. One dreaded and our problem we see that is one be a zero because there's no ex term here, and then C is one so that our problem b squared minus four a. C. It's just zero minus four, which is negative. That tells us that X squared plus one does not factor. So we must leave this in the denominator for the partial function. So at this point, we're taking this inter grant. Here we just rewrote this using our division, and now we go ahead and do partial fraction to composition. So the first turmoil here, non repeated and his linear. So we have a over X plus one. Since this quadratic did not factor, we're really checked. This is what the author calls case three. So we have B X plus C and then explore plus one. What's going on? Multiply both sides of this equation by the denominator on the left, and we'LL write the result in the next stage. So after multiplying by that denominator, we get for X on the list on the right, a X squared plus one, the explosive all times X plus one. And let's just go ahead and simplify this right hand side X squared. Plus they b X cleared CX dee eggs plus E. And then let's combine light terms a plus D X squared B plus C X, and then we have a policy at the very end. And now we just matched the corresponding terms on the left. We see that there is no X squared so we could right zero x square here. So we have zero. It was a plus B That's our first equation Now for the ex term. We see that it's a plus four on the left on the right, it's B plus C, so he must have B plus C equals for and the constants room on the left hand side of zero. And on the right hand side, it's a plus C. So we must have a plus. C equals zero. So this is a three by three system. We could go out and solve this for A B and C so I could rewrite the first equation by solving for a I could even take that last equation solvent for eh And then sense negative b a negative sear both equal. Say we must complete ease together to say b equals C And so plugging this into the second equation we have to be equals for So that means B is two. But that C equals Tuas Well, because C equals B and finally a equals negative scene. So that's negative too. So these are our values. So we'Ll go back to our partial fraction to composition will go ahead and plug in these values for a B and C and then we'll write out the corresponding in a girl. So the integral is now after plugging in our A, B and C, so that's from plunging the and A and then be was, too. And then see was also too. So that's two X plus two over X squared plus one. Let's go ahead and write. This is three separate rules, so negative two over X plus one. Pull out the two. Actually, it's not. Pull out to two. We have integral to X X squared plus one and then for the last time the girls good cloth and two one over Expert Plus one. And these are integral is that we can do for this first one here. If that plus one on the X is bothering you, feel free to do U substitution. So for this first integral, this should give us negative soon. Natural Log Absolute Value X plus one for the second inaugural new substitution u equals X squared plus one. Then do you equals two x t X. So this becomes in overall one over you, Do you? That's natural log absolute value. You and so that's Ellen of X squared plus one. Feel free to drop the absolute values here because X squared plus one is positive. And finally, for the last inaugural, you might remember this as the from the art and function. We know that the derivative of ten is one over X squared plus one. So it means that the integral of one over X squared plus one is ten inverse r and forgot my tanned by inverse on this hand. But if you forgot that fact, you can go ahead and feel free to do it. Use up here X equals one tan data. So using that tricks up here, this inner girl Oh, should be close should become two ten inverse x. So the last thing to do here is just at our three answers together for the intervals. So the first Indian girl we had negative too natural log, absolute value X plus one, Then the second integral on green natural log X squared plus one and then the last in a group over here on blue to ten in process. And don't forget the constancy of integration. And that's our final answer.