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JH
Numerade Educator

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Problem 56 Hard Difficulty

Evaluate
$$ \int \frac{1}{x^2 + k}\ dx $$
by considering several cases for the constant $ k $.

Answer

$\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}$

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Video Transcript

Let's evaluate this in un rolled by considering several cases for Okay, so here are three cases that will cover every single case out there. Case one K zero case to positive Kehr and case three negative here. So, in case one, we can rewrite this inner girl as just one over X cleared and then use the power rules evaluator Integral. So that would be the answer for Capel zero. Now, let's consider the case when K is positive. So then we have one over X squared, plus que we can rewrite that. His ex cleared plus radical case where and then we could use a trim stuff here. Here, we should take extra B. We recall radical care data and the ex Baruch. Okay, sequence where data and then this integral can be written as radical. Kay, she can't square. That's a K over here And then on bottom, When we square X, we get k times, tangents where and then plus this k here. So me got in fact throughout that care and then we use the fact that tan squared plus one is equal to C cancer. Pull off the constant And then now let's go to the next page. Since I'm going to get a room here, we have Groupe over. Okay, Times integral detail, which is just data. And then using our tricks of over here at sequels rookie attention. We could solve that for Dana. So divide both sides by the radical and then take the ark tan on both sides. Okay, so that will be our value for theta and then our constant of integration. See? So this would be our answer in the case when Kay is a positive number. And then now it's fine. We go to case three. This is one K is negative. So here we can rewrite this Integral s o. Basically, here we have a negative number. So the way to look at this for a simple example, like if we had X squared plus minus four. So if kay was a negative for you can write this in this form, and then you could do a difference of squares formula. So we can do this because Kate is negative. So in this case, it will be X plus root Negative, kay and then x some you go back. I got a little sloppy there. You have a negative can that radical and the negative. We put the negative inside a radical because then if K is negative, that means negative here cares positive d. X So this is what the author would call case one for the insta grant. We have distinct linear factors, so just looking at the inner grand Weaken break that up is a over. It's plus the first radical and then B and then we have the same, but with a minus. So let's go ahead and multiply both sides by this denominator on the left and you get one equals air X minus radical negative Kay E X plus radical negative. And then let's go ahead and pull out a X there. And then we left, though, are constant. It's a minus there. So on the left we see that there's no extra on the left. So over here, eh? Must be must be zero. The constant term on the left is one. So that means the constants from over here with no eggs, that must be one. So let me go ahead and write this on the next age. We have the room Negative, Kay, minus a route Negative Kay equals one. We also had a plus B equals zero. So solve the second equation and then go ahead and plug in B equals negative over here and then softer, eh? And now we have a and then using this, we find bee. And now we go ahead and plug in those A and B into our partial fraction of composition. And I were ready to integrate. So there's our a D. X and now we have our deal. Now, if this plus and minus thematic for bothering you feel free to do you some of this form for the first for the second. Very similar. In any case, when we go ahead and in every death, we should have you pull out that constants are negative. One over to root, negative can and then natural log Absolute value X plus radical negative pit. And then, for the second one, one over to negative kay Natural log X minus radical negative care. And of course, these terms can be combined using the long properties. And then we have our constancy of immigration. So this is our answer in case three, when K was negative