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.

35. ∫ x³/√(4x² + 16) dx

Accepted Answer

Hello. In this video, we're going to be computing the following indefinite integral. We are given the integral of X cubed divided by the square root of 5 X squared. Plus 25 DX. Now, we can approach this integral by applying a U substitution. We will allow you to equal to 5 X2 plus 25. If we have this U substitution, we must come up with a substitution for DX. In order to obtain the substitution, we will take the derivative of both sides of our U substitution, and we will get that DU is equal to 10XDX. And if we were to move the constant to the left hand side by dividing it, we will have 1/10 DU is equal to XDX. Now, there is a small error or small issue with this substitution. First of all, when we, if we look at the given integram, we have X cubed, but we only have XDX for our substitution, which means that we are left with two extra factors of X. Now, if we were to solve for X2 in our original U substitution, we can solve for X2, and that will give us X2 is equal to U minus 25 divided by 5. So in order to apply these substitutions, the first thing we're going to need to do is we're going to need to rewrite our our numerator. We can rewrite the numerator as a product of X2d and X, and moving the DX to the numerator will give us XDX for our DU substitution. So we will have X2d multiplied by XDX divided by the square root of 5 X2 + 25. And now, if we apply our substitutions, we will be left with the integral of the quantity U minus 25 divided by 5, multiplied by 1/10. Do you and this will be divided by the square root of you which can be rewritten as you to the half power. Now, what we're going to do is we're going to move some constants around. We know that in the numerator, if we multiply 5 and 10 together, we get the value of 50, and moving that to the front of the integral is going to give us 1 divided by 50. So we will have one divided by 50, multiplied by the quantity U minus 25 divided by U to the half power, DU. We can further simplify this by using algebraic methods. If we break apart the integrand as a difference of fractions, we will have 1 divided by 50, multiplied by the integral of U divided by U to the half power. 25 divided by you to the half power. And if we were to express these as a single term of you, we will have 1 divided by 50 multiplied by the integral of u to the power minus 25 multiplied by u to the negative half power tou. Now, what we can do is we can go ahead and take the antiderivative of each of the terms of the integral. That is going to give us 1 divided by 50, multiplied by the quantity, 2/3 multiplied by U to the three halves power, minus 25 multiplied by 2, multiplied by U raise to the half power, and because we are solving for a general integral, we will add a general constant C. Now, if we multiply the coefficients together, we can further simplify this to be 1 divided by 75, multiplied by U to the three halves power, minus U/ power plus C. At this point, we are going to resubstitute you back in the terms of X. Now originally, we said that U is equal to 5 X2 + 25. So by resubstituting these values, we will get 1 divided by 75, multiplied by the quantity, 5 X2 + 25, raised to the 3 halves power. Minus the quantity, 5 X2 + 25, raise to the half power plus C. Now, here, what we are going to do is we are going to factor out a common term. Let's go ahead and factor out 1 divided by 75 and 5 X2 plus 25 raised to the half power. By factoring out these terms, we will have 1 divided by 75, multiplied by 5 X2, plus 25 raised to the half power, multiplied by the quantity, 5 X squad. Plus 25. Minus 75 plus C. By combining like terms together, we will have 1 divided by 75, multiplied by 5 x 2 plus 25, raised to the half power, multiplied by the quantity, 5X2 minus 50. Plus C And once we reexpress the half power as a square root, we will get the simplest form of the solution, which is 1 divided by 75, multiplied by the square root of 5 X2 plus 25 multiplied by the quantity. 5 x 2 minus 50 plus C. One more thing we can do here, if we were to factor out a 5 from 5 X2 minus 50 and multiply that to 1 divided by 75, we will end up with 1 divided by 15, multiplied by the square root of 5 X2 + 25, multiplied by the quantity, X2 minus 10 plus C. This is going to be the simplest form of the solution to the given indefinite integral, and with that being said, the overall solution is going to be a. 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.
35. ∫ x³/√(4x² + 16) dx

Key Concepts

Integration Techniques

Substitution Method

Rational Functions and Square Roots

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