Evaluate the indefinite integral.∫x(x−6)5dx\int_{}^{}\frac{x}{\left(x-6\right)^5}dx∫(x−6)5xdx

− 1 3 ( x − 6 ) 3 − 3 2 ( x − 6 ) 4 + C -\frac{1}{3\left(x-6\right)^3}-\frac{3}{2\left(x-6\right)^4}+C

Evaluate the indefinite integral. ∫ x ( x − 6 ) 5 d x \int_{}^{}\frac{x}{\left(x-6\right)^5}dx ∫ ( x − 6 ) 5 x d x

this problem, we want to evaluate the indefinite integral of x over x minus six to the power of five dx. So let's go ahead and jump right into our steps for substitution here by choosing u and finding du. Now looking at this problem, I see this x minus six inside of parentheses raised to the power of five. So that's a good clue that this is what I should choose as my u. So if u is equal to x minus six, what is du? Well, du is the derivative of x minus six times d x. The derivative of x minus six is just one. One times d x is just d x. So here du is just equal to d x. So I have my u. I found du. We can move on to step two, rewriting our integral only in terms of this u and du. So looking at my integral here, that x minus six is u, that dx is my du, but I can see that I have this extra variable here. Now whenever we have this extra variable, remember that we can rewrite it in terms of u in order to make our substitution work. So if u is equal to x minus six, if I go ahead and just add six to both sides here, that means that x is then just equal to u plus six. So with that in mind, I did go ahead and rewrite my integral because this x is just u plus six. So rewriting this, this is the integral of u plus six over u to the power of five du. So we've fully rewritten this integral only in terms of u and du, and we can move on to step three, actually integrating this with respect to u. Now we can do some splitting off of this fraction in order to make this easier to work with here. Because we have those two terms on the top and just one term on the bottom, we can split them off keeping that denominator the same. So this is u over u to the power of five and then plus six over u to the power of five, just splitting that fraction off into two separate fractions. And integrating here with respect to u, doing a little bit further simplification before we get to that antiderivative part here, u to the power of one over u to the power of five. If I take that one subtract that five, that's gonna be u to the power of negative four. Then I'm gonna go ahead and rewrite this just not as a fraction at six times u to the power of negative five du. So now we can go ahead and just use the power rule here. We're going to add one to our exponent, divide by that new exponent. Negative four plus one is going to be negative three, so this is negative one third times u to the power of negative three and then a plus six times, applying the power rule here, negative five plus one, and then dividing by that negative five plus one as well, that's gonna be negative one fourth, u to the power of negative four and then plus my constant of integration c because this is an indefinite integral. Now we can do some cleaning up here. This is negative one third u to the power of negative three and then combining that all into one fraction putting that negative out front, that's negative. Six over four is going to simplify to three over two. So three over two u to the power of negative four plus c. Then we can move on to our final step here replacing u with our original function of x. We know that that is x minus six, so we wanna replace both of these u's with that. So this is negative one third times x minus six to the power of negative three minus three halves times x minus six, replacing our u to the power of negative four plus c. Now because we have these negative exponents here, we could go ahead and flip these to the bottom of our fraction. But if you wanna write it this way, this is still correct. I will go ahead and show you if we wanna take that approach and make all of our exponents positive. This would be one over three times x minus six to the power of three, just on the bottom there, and then minus three over two times x minus six to the power of four plus that constant of integration c. Either of these ways of writing this answer are correct. Feel free to let us know if you have any questions here and I'll see you in the next one.

Evaluate the indefinite integral. ∫ x ( x − 6 ) 5 d x \int_{}^{}\frac{x}{\left(x-6\right)^5}dx ∫ ( x − 6 ) 5 x d x

− 1 3 ( x − 6 ) 3 − 3 2 ( x − 6 ) 4 + C -\frac{1}{3\left(x-6\right)^3}-\frac{3}{2\left(x-6\right)^4}+C

− x 6 ( x − 6 ) 6 − 3 x 6 ( x − 6 ) 6 + C -\frac{x}{6\left(x-6\right)^6}-\frac{3x}{6\left(x-6\right)^6}+C

− 1 3 ( x − 6 ) 6 − 3 2 ( x − 6 ) 6 + C -\frac{1}{3\left(x-6\right)^6}-\frac{3}{2\left(x-6\right)^6}+C

1 3 ( x − 6 ) 3 + 3 2 ( x − 6 ) 4 + C \frac{1}{3\left(x-6\right)^3}+\frac{3}{2\left(x-6\right)^4}+C

8. Definite Integrals

Substitution for Indefinite Integrals

Business Calculus

8. Definite Integrals

Substitution for Indefinite Integrals

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{}\frac{x}{\left(x-6\right)^5}dx$

$- \frac{x}{6 {(x - 6)}^{6}} - \frac{3 x}{6 {(x - 6)}^{6}} + C$

$- \frac{1}{3 {(x - 6)}^{6}} - \frac{3}{2 {(x - 6)}^{6}} + C$

$- \frac{1}{3 {(x - 6)}^{3}} - \frac{3}{2 {(x - 6)}^{4}} + C$

$\frac{1}{3 {(x - 6)}^{3}} + \frac{3}{2 {(x - 6)}^{4}} + C$

Verified step by step guidance

Step 1: Rewrite the given integral in a more manageable form. The integral is ∫x(x−6)^5 dx. Expand the integrand by distributing x into (x−6)^5, or consider substitution if it simplifies the process.

Step 2: Use substitution to simplify the integral. Let u = x−6, which implies du = dx and x = u + 6. Substitute these into the integral to rewrite it in terms of u.

Step 3: After substitution, the integral becomes ∫(u+6)u^5 du. Expand the expression (u+6)u^5 into u^6 + 6u^5 and split the integral into two parts: ∫u^6 du + ∫6u^5 du.

Step 4: Apply the power rule for integration to each term. The power rule states that ∫u^n du = (u^(n+1))/(n+1) + C, where n ≠ -1. Integrate each term separately.

Step 5: After integrating, substitute back u = x−6 to return to the original variable x. Combine the constants and simplify the expression to match the given answer format.

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