Evaluate the indefinite integral.∫1(3x+2)5dx\int_{}^{}\frac{1}{\left(3x+2\right)^5}dx

− 1 12 ( 3 x + 2 ) 4 + C -\frac{1}{12\left(3x+2\right)^4}+C

Evaluate the indefinite integral. ∫ 1 ( 3 x + 2 ) 5 d x \int_{}^{}\frac{1}{\left(3x+2\right)^5}dx

we're asked to evaluate the indefinite integral of one over three x plus two to the power of five d x. Now we're gonna do this using substitution. So let's go ahead and jump into our first step here where we wanna choose u and find du. Now remember that u should often be an inside function in parentheses under a radical or something like that. And here we can see that we really only have one possible choice for you, this three x plus two in our parentheses here. So if I make u equal to that three x plus two, what does that make du? This is gonna be the derivative of that three x plus two times d x. So the derivative of three x plus two is just three, multiply that by d x, we have three d x. So with step one done, we move on to step two, rewriting our integral only in terms of this u and d u. So if this three x plus two is our u, d u is at three times d x. We don't have that three in our integral integral currently. So we need to go ahead and multiply by that missing constant and its reciprocal in order to make this substitution work. So if I multiply by three in here, then I have that three d x, which is my d u. I also need to go ahead and multiply by that reciprocal which I'm gonna go ahead and just put on the outside of my integral here. So that means that I can rewrite this as one third times the integral of one over u to the power of five du. So now that that's rewritten in terms of u and du, we can move on to step three, integrating with respect to u. Now I'm gonna go ahead and rewrite this one over u to the power of five as u to the power of negative five just so it's more clear how we're gonna use the power rule here here to get this antiderivative. So I wanna go ahead and increase that power by one and then divide this by that new power negative five plus one. So that will give me here a one third keeping that original constant And then since I'm dividing by a negative five plus one, that's gonna be a negative one fourth, u to the power of negative four, and then plus my constant of integration c. Now we're gonna go ahead and multiply these two constants together in order to simplify. This is gonna be negative one twelfth times u to the power of negative four plus that constant of integration c. So with that integral found, we wanna go ahead and complete our final step here, which is replacing u with our original function of x. Remember that original thing we chose for u was three x plus two, so that's what we're gonna plug in for u here. So this is negative one twelfth times three x plus two to the power of negative four plus that constant of integration c. Now because we have this negative power here, we can rewrite this in a way that we only have a positive power by putting it on the bottom of our fraction. So this would be negative one over 12 times three x plus two to the power of positive four plus that constant of integration c, and this gives us our final simplified answer. Let us know if you have any questions, and let's keep practicing.

Evaluate the indefinite integral. ∫ 1 ( 3 x + 2 ) 5 d x \int_{}^{}\frac{1}{\left(3x+2\right)^5}dx

− 1 12 ( 3 x + 2 ) 4 + C -\frac{1}{12\left(3x+2\right)^4}+C

− 1 5 ( 3 x + 2 ) 4 + C -\frac{1}{5\left(3x+2\right)^4}+C

1 12 ( 3 x + 2 ) 4 + C \frac{1}{12\left(3x+2\right)^4}+C

1 5 ( 3 x + 2 ) 4 + C \frac{1}{5\left(3x+2\right)^4}+C

8. Definite Integrals

Substitution

Calculus

8. Definite Integrals

Substitution

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{} \frac{1}{{(3 x + 2)}^{5}} d x$

$- \frac{1}{12 {(3 x + 2)}^{4}} + C$

$- \frac{1}{5 {(3 x + 2)}^{4}} + C$

$\frac{1}{12 {(3 x + 2)}^{4}} + C$

$\frac{1}{5 {(3 x + 2)}^{4}} + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a rational function of the form $ \frac{1}{(3x+2)^5} $. To solve this, we can use substitution. Let $ u = 3x + 2 $, which simplifies the denominator.

Step 2: Differentiate $ u = 3x + 2 $ to find $ du $. This gives $ du = 3dx $, or equivalently, $ dx = \frac{du}{3} $. Substitute $ u $ and $ dx $ into the integral.

Step 3: After substitution, the integral becomes $ \int \frac{1}{u^5} \cdot \frac{1}{3} du $, which simplifies to $ \frac{1}{3} \int u^{-5} du $.

Step 4: Apply the power rule for integration, which states $ \int u^n du = \frac{u^{n+1}}{n+1} + C $ for $ n \neq -1 $. Here, $ n = -5 $, so the integral becomes $ \frac{1}{3} \cdot \frac{u^{-4}}{-4} + C $.

Step 5: Simplify the result to $ -\frac{1}{12} u^{-4} + C $. Finally, substitute back $ u = 3x + 2 $ to return to the original variable, giving $ -\frac{1}{12(3x+2)^4} + C $.

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