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JH
Numerade Educator

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Problem 32 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int_0^1 \frac{x}{x^2 + 4x + 13}\ dx $

Answer

$$
\frac{1}{2} \ln \frac{18}{13}-\frac{\pi}{6}+\frac{2}{3} \tan ^{-1} \frac{2}{3}
$$

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Video Transcript

Let's evaluate the given in several here to see if we could do partial fraction the composition. The first thing we should do is to see if that denominator factors So we'LL look at X square clothes for X Plus thirteen here was going to write that one in the front that's R A. We see their beers core and then C is thirteen. So to see if this factors let's just check the discriminative B Square minus four a. C. So what we get here be square sixteen minus four times one times thirteen. That's a negative number, So that means that this quadratic won't factor. So in that case, this fraction already is a partial fraction to composition. There's no more work to do there, but to integrate. We should go ahead and try as some kind of substitution. You sub may not work so easily here, so let's go ahead and try toe, take that denominator and complete the square. So we have four ex half a forest to the score, that unit for and then thirteen minus four so we can write. The denominator is X plus two squared plus nine. We're just three squares, so coming back over here. X plus two Cube, six square, blustery swear. Excuse me. There we go. And then d X Now let's go ahead and try it tricks up. Looking at the denominator. We should take X Plus two to be three ten data, then taking the differential on the side. The differential derivative of two zero. So there's nothing else to write on the left and on the right. Three seconds were their leader also looking at the denominator. We can rewrite this as well, so let's go ahead. Since we're a new variable data, let's just go ahead and leave these unknown. Let's just call it Data one and data, too. So on the tab, we see that there's a X. So how do we get X? We get that from solving this equation for X, so subtract two on the side. X equals three ten Tatum minus two. So that corresponds to the X here. Then we have DX. But we found that that's three seek and square data, Dana. And then on the bottom, we see X plus two squared. So that's just three Tanja and we'LL Square that whole thing. That's nine time ten Square plus three squared, which is nine. So let me just go. In fact, at that nine now we could use the pathetically an identity that tan squared plus one. Two c can't square. So these will cancel. And then let's just go ahead and read everything else carefully dated one they did, too. Three ten data minus two and then three over nine is just We have a three on the bottom that's over Now. We could go ahead and split this into two fractions, and then we have in this case, so the threes cancel. So that's just the tan. And then here we just have to over three. So just just simplifying the fractions and then the next page will start integrating. So the first integral in a room with him for that if you don't remember how to do this, one are just memorized the formula. You could write to him a sign of Rick or sign and then go ahead and let you be co sign and doing use up. So going to the next page, we have natural log, absolute value seeking. So that was for the integral tan and then minus two over three data our end point state a one data to and then this is where we draw the triangle to get back into our original limits. So we do not have to use data one and data too. The strong right triangle two tricks, um, that we had if he divided by three. So that tells you that the adjacent over Excuse me the opposite over the adjacent is X plus two over three. And then using put that grand era, you could find the high powers if you want, you could even go ahead and rewrite the hypothalamus. I just multiplying things. Oh, and you actually see the original denominator. It's still there. But this time inside the radical. So now let's just evaluate c can. So we have natural log absolute value, So C can will be high pond news over adjacent. So it's just a radical divided by three. And then we have to over three and then theta. You could find data by just taking the tricks of up here and taking art e'en on both sides. There's data, so we come over here and plug that in for theta, so it's hand in verse, expose to over three and then now we're back into X so we could use our original limits zero and one. So now the last time, Just plug in these numbers and simplify as much as we can. So first we plug in one we'LL plug it in for X so we'll have one and then four that's five and then add that's a thirteen. We have a radical eighteen up there and then also plug in one for X over here. So you have to over three. Is he an inverse of one? And then now we plug in the zero when we played in the zero for X will just have a radical thirteen up there and then we'LL have minus two over three ten inverse of two thirds. And then the last thing we could do here is maybe clean this expression up a little bit. We know ten inverse of one. It's pie. Before now we can go ahead in use the log properties here to rewrite these expression. So one way to go here is too. We see that you're dividing there and we know that natural idea. A over B is Ellen Day minus Ellen be so you can use that fact to rewrite this. Live with them. You could do the same thing for the second longer of them here. You know, when you do so that will ln threes will cancel out. So you have Ellen radical eighteen. And then we have minus Ellen Radical thirteen. And then now we have We had power before here. So we have to over three every floor and so that'LL just be pi over six and then watch out for this double minus here. That's a plus to over three arc Tan also to over three. And there's a final answer.