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JH
Numerade Educator

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Problem 23 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{dx}{\sqrt{x^2 + 2x + 5}} $

Answer

$\ln \left|\sqrt{(x+1)^{2}+4}+x+1\right|+D$

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Video Transcript

here we have the integral of DX over the switcheroo of X squared plus two x plus five. So let's go ahead and take this quadratic and complete the square. So half of two is one. And if I square that I also get one. So let's put a plus one. And then, since we added one we'LL make up for by subtracting one no. So let's factor this term in apprentices. That's X plus one square plus four. So if we want, we could write This is too square now, ideally for a truths of we would prefer something of the form ex cleared plus a square. But we don't quite have that because we have this plus one. So if that plus one is bothering you, let's do it use up, take you to be X plus one. Then do you is just the X so we could rewrite our general as do you over and actually say one more thing here. After this use institution, we can rewrite the quadratic is useful here plus two squared so we could replace this quadratic and X with you squared plus two squared. So that's a no go inside the radical. And now we have a new integral that's tailor made for tricks up here. We could take, you know, be tooth and data then do you is to seek and squared later. So let's take this expression and the radical So replacing you with two ten data. We have four chance for data plus four. That's a fact There on the floor. Then we have tan squared plus one so radical forest too. And we have the square room of sequence where data about the protagonist identity and taking the square root we have to seek and data so we could rewrite are integral. So do you now becomes to seek and data to seek and square debater and the radical which we just simplified down here in the bottom, right, that's too thick and NATO canceling the twos. One of the sea cans we're left over with integral of C can't and we know this in a rule to be the natural log Absolute value seek and data plus tan data plus E Now I'm running on room here. Let me go to the next stage. This was our tricks up. Maybe rewrite this and the integral that we just left off with was natural log seeking data plus Dan Data. Plus he now going to our tricks up. We can rewrite this as tan data equals you over, too. So let's draw a triangle. We know that tension is opposite over adjacent, since that equals you over to Let's just go ahead and take Decide to be you and decide to be too. What's the note? The high pardons by age. Bye, Petya Green Serum a squared is you squared plus two squared. So take the square root and we get H. Now we can evaluate, seek and as well so she can is high powered news over the adjacent, So each other, too. And then tangent is just you over, too, so we can rewrite this so there's two steps we can apply here. We couldn't first replace you with X plus one that was our U sub in the very beginning, So we have X plus one squared plus four and let's combine these fractions because they have to have the same denominator plus you, which is X plus one all over, too. And then we could go ahead and use one of the properties for logarithms. That says Ellen, Eh, Overby is Ellen a minus? Ellen, be so here. B is two. So using a lot of property, we can write this. And then if we like, we can go ahead and replace the C minus Ellen to with another constant D. And there's our final answer.