2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

74. ∫ dx/√(√(1 + √x))

Accepted Answer

Hello. In this video we are going to be solving for the given indefinite integral. We are given the integral of D X divided by the square root of the square root of 3 plus the square root of X. So in order to approach this problem, let's go ahead and first algebraically simplify this given embedded square root. Now, we can represent the first two square roots as the exponent of some quantity raised to the half power. So we can rewrite the denominator to be the quantity 3 plus the square root of X. Raised to the half power which this quantity itself is raised to the half power and by multiplying the exponents together, we can rewrite this as 3 plus the square root of X raised to the 1/4th power. Now this is going to be located in the denominator once we simplify. But if we were to represent this value with a negative exponent, we can algebraically simplify the integral to be the integral of the quantity 3 plus the square root of X raised to the -14 DX. Now, in order to further simplify this problem, we are going to need to use a U substitution. We are going to allow you to equal to the square root of X. Now, this U substitution won't help us that much, because we need to find another substitution for DX. So let's go ahead and sell for X in this case. We are going to square both sides, and that will give us X equal to U2. Now, by taking the derivative with respect to X, we will end up with D X equal to 2UDU. So, we have a U substitution for the square root of X and a substitution for DX. That is going to allow us to rewrite this integral to be the integral of 3 + U raised to the negative 1/4 power multiplied by 2 U. Do you Now, one thing we can do here is we can move this 2 to the front of the integral as a constant. And now what we want to do is we want to apply one more substitution to further simplify the integral. Let's go ahead and apply a substitution using w. We are going to allow W to equal to 3 plus U, and by taking the derivative with respect to W on both sides of this substitution, we will get DW is equal to DU. But now we need to find a substitution for this extra U term. If we were to solve for you in our W substitution, we will get that U is equal to W minus 3. And so if we apply this substitution, we can further rewrite the integral to be 2 multiplied by the integral of sorry, W. Raised to the 1/4th power. Multiplied by the quantity, W minus 3DW. Now what we are going to do is we are going to solve by further using algebra. We will distribute to the negative 4th power to each of the terms in W minus 3, leaving us with 2 multiplied by the integral of W 3/4 minus 3W to the -14 DW. And by taking the antiderivative of each term within the integral, we will be left with 2 multiplied by 4/7, multiplied by W to the 7/4 power, minus 3 multiplied by 4/3, multiplied by W raised to the 3/4 power. And because we were working with a general integral, we will add a general constant C. Now, if we were to multiply the coefficients together, we can further simplify this to B 8/7 multiplied by W raises the 7/4 power, minus 4 multiplied by W raised to the 3/4 plus C. Now what we want to do is you want to go ahead and rewrite this from W back in terms of you. Now, originally we let W equal to 3 + U, so reapplying the substitution is going to give us 8/7 multiplied by the quantity, 3 + U raised to the 7/4 minus 4 multiplied by the quantity, 3 plus U raised to the 3/4 plus C. Furthermore, we want to rewrite this U term back into terms of X, and our original U substitution was U is equal to the square root of X. So by reapplying the substitution, we will be left with 8/7 multiplied by 3 plus the square root of X raised to the 7/4s power. -4 multiplied by 3 plus the square root of X, raised to the 3/4 plus C. And this is going to be the simplest form of the antiderivative for the given indefinite integral. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
74. ∫ dx/√(√(1 + √x))

Key Concepts

Integration Techniques

Substitution Method

Radical Functions

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