7–40. Table look-up integrals Use a table of integrals to eval...

Question

7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
34. ∫ dx / (x(x¹⁰ + 1))

34. ∫ dx / (x(x¹⁰ + 1))

Accepted Answer

Welcome back, everyone. Evaluate the indefinite integral. Integral of DX divided by X multiplied by X to the power of 6 + 1. For this problem we're going to use U substitution. Let's suppose that U is equal to X the power of sex. Then the derivative DU divided by DX is 6 X to the power of 5, right? We're using the power rule. And now what we can do is simply sold for the X, right? So if we express the X, then the X equals the U divided by 6 X to the power of 5. Now, what we want to do from here is basically analyze our integrals, so that's integral of DX divided by X. Which is multiplied by X to the power of 6 + 1. And we can substitute for the X to begin with, which is the EU. Divided by 6 acts to the power of 5. And we are dividing by X. So basically 6 X the power of 5 is multiplied by X. And then we're multiplying by X to the power of 6 + 1, right? So that would be U + 1. Now notice that 6 X of the power of 5 multiplied by axis, 6 X of the power of 6. So we can just write 6 X to the power of 6, or basically 6 U, right? Because now we have X to the power of 6, which is U. Well then, so now we are integrating DU divided by 6, U multiplied by U + 1. We can factor out the constant 1/16. Integral DU divided by. U which is multiplied by u + 1, and because we have a product of two linear functions in the denominator, we're going to use partial fraction decomposition. So let's consider our fraction within the integral that would be 1 divided by U, which is multiplied by U + 1. We can write it as a sum of two fractions, a divided by U plus. B divided by U + 1. So we can now find the common denominator. We already know what it is, right? That's U multiplied by U + 1, which means that A is multiplied by You plus one. And B is multiplied by U. So in the numerator we get a multiplied by U + 1 + B U. We can now distribute which gives us a U + A + BU. Divided by U which is multiplied by U plus 1. And then we can combine the like terms. So, the terms that contain you are a U and BU, so we can write it as A plus B. Multiplied by you. And then we have the three coefficients we're adding a. This is divided by U which is multiplied by U + 1. And now looking at this form, we can first of all say that A plus B. is equal to 0 because the original fraction does not contain any u terms on the numerator and we can also say that A is equal to 1. Because the only free coefficient that we have is one in the original fraction. So, from the first equation, B is equal to negative A, meaning B is equal to -1. And therefore we can now rewrite our integral, which is 1/6 integral of a sum of two fractions. So a divided by U, which is 1 divided by U. Plus B divided by U + 1, right? So since B is -1, we are just subtracting 1 divided by U + 1. Do you And then we have to recognize that we have linear functions in the denominator, right? In each case, our slope is once we can basically apply. The formula for a basic integral of a 1 divided by U, right, which is LN of the absolute value of U. So, the first integral would be LN of the absolute value of U, and then we are subtracting LN of the absolute value of U + 1. And let's add a constant of integration C and simplify so that be 1/6 Ln of the absolute value. Now because we're subtracting two logarithms with the same base of e, we can just divide u by U + 1 within the same log using the properties of logarithms. So since U is X the power of 6, we get a n of the absolute value of X the power of 6. Divided by X to the power of 6 + 1 and finally plus C, so that'll be our final answer for this problem. Thank you for watching.

7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
34. ∫ dx / (x(x¹⁰ + 1))

Key Concepts

Indefinite Integrals

Integration Techniques

Tables of Integrals

Watch next

7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.34. ∫ dx / (x(x¹⁰ + 1))

Key Concepts

Indefinite Integrals

Integration Techniques

Tables of Integrals

Watch next

7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
34. ∫ dx / (x(x¹⁰ + 1))