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Numerade Educator



Problem 46 Hard Difficulty

Make a substitution to express the integrand as a rational function and then evaluate the integral.

$ \displaystyle \int \frac{\sqrt{1 + \sqrt{x}}}{x}\ dx $


$4 \sqrt{1+\sqrt{x}}+2 \ln |-1+\sqrt{1+\sqrt{x}}|-2 \ln |1+\sqrt{1+\sqrt{x}}|+C$


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Video Transcript

let's make a substitution to evaluate the integral. Let's try you to be the numerator That's one plus radical X all inside the square root And before we differentiate, let's just square both sides Get rid of one of those radicals and then we can differentiate each side, use the power rule And then now we can go ahead and write this in terms of you. So Rule X, we'll just take this equation and solve for Rue X. Yeah, and then multiply both sides of this equation here. Bye the denominator. And then you get for you U squared minus one. Do you equals D X? So that will replace D X and the integral. So we have integral on top. We see that radical up here in the numerator. That's just you, by definition. So that's easy That just becomes you. But then DX more complicated expression. We just found this on the side and that's for you. And then you square minus one do you all over X and X. Well, we just get that by squaring both sides of this equation so we just have you squared minus one squared, mhm. And here we should cancel we see that we have you squared minus one on top and bottom so we can cross that off. We just have one left on bottom. We can also take out that for so we're left over with U squared minus one. So it's right that no need to factor yet because here the numerator has same degree as the denominator. They're both too. So we should do long division here. So let me go to the side to do that u squared over u squared as one multiply, subtract, we get a remainder of one. This means that we can write the integral as the quotient, which was one plus one from the remainder divided by the original denominator. Now we have this integral to deal with. You may memorize this from a table, but if not, there's no problem because we can write it is one over you plus one and then you minus one. And then we could do partial fraction decomposition using what the author calls case one distinct, non repeating linear factors. So let's go to multiply both sides of this equation here by the denominator on the left. So the left side just becomes one. The right side becomes a you plus one B U minus one. Now I'm running out of room here. So let me go to the next page. Yeah, one equals factor out of you. And then our constant term left over. And we see, since there's no you on the left, we see that a plus b must be zero. And we see that a minus B must be one, because that's the constant term on the left. So we have a two by two system. Okay, let's go ahead and add those together we get to eight equals one, right? That means a is a half and therefore be is also be will be negative one house by using the second equation here. Mhm. So going back to our integral. Remember, we pulled out the floor. We are one. And now we originally had one over U squared minus one here, but then we just did partial fraction decomposition. There's are a over U minus one, and then negative one half over you plus one deal. Now we're ready to integrate. If these minus one and plus one are bothering you. Feel free to do a use up here for the second one you can do w equals U minus one. For this third inaugural, you can do you plus one. In either case, when we integrate, we should get we have for you to natural log absolute value U minus one Negative two natural log. Absolute value. You plus one. Don't forget the constant seed and then finally use the original definition of you. The substitution to go ahead and replace you in terms of X. So four and then you, by definition one plus radical X inside the radical and then to natural log. Then we have absolute value. Yeah, technically, you can drop the absolute value here if you write it this way, because this first radical is bigger than one. But that's not too important right now. And then we have you plus one absolute value and then plus our constancy of immigration. And that will be our final answer. Mhm.