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JH
Numerade Educator

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Problem 69 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int_1^{\sqrt{3}} \frac{\sqrt{1 + x^2}}{x^2}\ dx $

Answer

$$\sqrt{2}-\frac{2}{\sqrt{3}}+\ln (2+\sqrt{3})-\ln (1+\sqrt{2})$$

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Video Transcript

Let's use integration by parts for this one. Let's go ahead and take you to be the Numerator Explorer plus one inside the radical. Then do you go ahead and use the chain rule here from Kokkalis and then simplify should get this for do you and then we're left over with DV equals less right as X to the minus two and this gives us that the so here you have to integrate to use the power rule. Negative one over X. Yeah, so they're recalled the formula UV minus integral video. So let's go ahead and plug in Are you and being and let's not forget the limits here. This is a definite our world's on our problem. We will still be using these limits here. So we have you times be that will give us That's a negative in front of the radical oops over X And then we plug in one and then route three. So that's are you. Times be there this term over here and then now will do the subtraction And then we have this term over here, so we have negative and then we have inner girl and then V and then we have, do you as well. So let's go ahead and write that. So VD you that gives us negative one over X and then X over square root, X squared plus one. And here we see that we can cancel these minuses that gives us a plus. And then we could cancel those exes as well. And we just have the integral of one over this square. Next plus one, you can use a trick. So for this one, go ahead and take X to be tan data to evaluate the tricks up there. And let's not forget our limits. One blue three still, just as the original problem. So here I go, after using that tricks up and then for all so we could plug in these limits into this expression. So plug in room three first and then because of this minus sign here, we add that's plugging in one right there and then after in aerating this thing, using the tricks of and then going back in terms of X, this's our anti derivative. And then we still have the same end points after sense for back in X. So it's going a lot Next page here. The last thing to do is this toe plug in thes two values for X into this expression here, and that's a threat. And so, after simplifying we have room to that's from the root to over one square before is too. So that's negative to O. R. Room three and then natural Log to plus through three and then minus natural log one plus square or two. And we do not need absolute value here because these terms are both positive and that's a finalist.