Evaluate the integrals in Exercises 39–54.
∫ 1 / ((x¹/³ ...

Question

Evaluate the integrals in Exercises 39–54.

∫ 1 / ((x¹/³ - 1)√x) dx

(Hint: Let x = u⁶.)

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral by using the given substitution. The integral given to us is the integral. Of 1 divided by the quantity x raised to the power of 1/3 minus 4 multiplied by the square root of x d x. Now the substitution that the problem wants us to use is x is equal to u raised the power of 6. So since we are substituting X in terms of U, we have to change everything in our integrand from terms of X to terms of U. So we need to come up with a substitution for DX. We will take the derivative of this X replacement, and we will get DX is equal to 6 multiplied by U raised to the 5th DU. So, this is going to be our substitution for X. So the way we apply this substitution is that everywhere that we see an X term, that is going to be replaced with U to the 6th, and DX is going to be replaced with 6 U raised to the 5th DU. So by applying our substitution, we can rewrite the integral to be 6U raised to the 5th divided by the quantity U 2 minus 4 multiplied by U raised to the 3rd DU. Here we can eliminate 3 multiples of u from both the numerator and denominator, and this will allow us to rewrite the integral to be the integral of 6 u 2 divided by u2 minus 4DU. Now here we can further simplify the integral by using an addition technique. We will introduce a -4 plus 4 in the numerator, and we will group up the three terms together. This will allow us to rewrite the integral into the sum of two integrals. We will have the integral of 6 multiplied by u2 minus 4, divided by u2 minus 4. And then we will add the integral of 24 divided by u square minus 4. Now the first integral will further simplify to just be 6, so we are left with the integral of the constant 6 plus the integral of 24 divided by u square minus 4. So, let's go ahead and label the first integral as integral 1, and the second integral as integral 2. And what we're going to do is we're going to evaluate each of the integrals independently. Now integral one is fairly simple to integrate because the integral of a constant. With respect to you, it's just going to be 6 you. So that's going to be the value of our first integral. But the second integral is a little bit trickier. If we are taking a look at the 2nd integral, we can factor the denominator by using a difference of squares. So we can rewrite the integral to be the integral of 24 divided by U. U minus 2 Multiplied by u plus 2, and we can simplify this integral by using partial fraction decomposition. So we will go ahead and state the partial fractions. Now, each of our multiples in the denominator is a linear function of U, so our partial fractions are going to be A divided by U minus 2, plus B divided by U + 2. Now, if we were to multiply this entire equation by the common denominator, we can further rewrite the equation to be 24 is equal to A multiplied by U + 2 + B multiplied by U minus 2. Now multiplying and gathering terms together allows us to rewrite the equation as 24 multiplied by A plus B multiplied by U plus the constant terms 2A minus 2B. This will give us the following system of equations by setting the coefficients equal to each other. So by setting the coefficients of U equal to each other on both sides of the equation, we will get A + B equal to 0. And by setting the coefficient constants equal to each other, we get 2A minus 2B is equal to 24. Now, once we solve this system, we will get the coefficients A is equal to 6 and B is equal to -6. So by plugging this back into our partial fractions, we will get the partial fractions 6 divided by U minus 2 minus 6 divided by U + 2. So we can split the integrand into a difference of two fractions, and by assigning each of them their own integral symbol, we can rewrite the overall integral to be the integral of 6 divided by u minus 2 minus the integral of 6 divided by u plus 2. Now, here, we can use our natural logarithm identity to integrate both of these integrals. So they are going to integrate to be 6 multiplied by the natural logarithm of U minus 2 minus 6 multiplied by the natural logarithm of U + 2. And by using our difference property of the natural logarithm, we can further combine these to be 6 multiplied by the natural logarithm of U minus 2 divided by U plus 2, and that is going to be the result of the second integral. So if we go back and combine the two results together, the overall antiderivative is going to be 6U. Plus 6 multiplied by the natural logarithm. Of u minus 2 divided by u plus 2. Now recall from earlier, 01 more thing I forgot, we have to add a generic constant C because this is an undefined integral. But remember earlier in the problem that we were asked to use the substitution X is equal to U to the power of 6. Well, now that we have the antiderivative in terms of u, we have to rewrite it back into terms of x. So if we were to solve u in terms of x for the substitution, u is going to equal to x raised to the power of 1/6. And so now if we replace every u with x to the power of 1/6 in our denominator, our final answer is going to be 6 multiplied by x raised to the power of 1/6 plus 6 multiplied by the natural logarithm of x raised to the power of 1/6 minus 2. Divided by x raises the power of 1/6 plus 2. Then we will add the generic constant, and this is going to be the simplest form of the solution to the general integral. And with that being said, the overall solution to this problem is going to be option B. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Evaluate the integrals in Exercises 39–54.∫ 1 / ((x¹/³ - 1)√x) dx(Hint: Let x = u⁶.)

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Evaluate the integrals in Exercises 39–54.
∫ 1 / ((x¹/³ - 1)√x) dx
(Hint: Let x = u⁶.)