Trouble with this concept regarding u and antiderivatives.

[calc67x]

I have a problem like this : x(2x+1)^1/3
We need to find the antiderivative.
The step I don't understand, which is in the book, is this:
du=(2x+1)^1/3
That would be 2/3(2x+1)^2/3 because u=(2x+1)^1/3
That makes sense to me, as it is the derivative of (2x+1)^1/3.
Then we have x=u³/2-1/2, which makes sense.
However, I then see where it says:
x(2x+1)^1/3= 3/2 u³(u³/2-1/2)The only part that does not make sense to me is how did they get the 3/2u³?
I would think that derivative of u³/2-1/2 would be 3/2u²
After this step, the rest of the problem is easily understood.
The answer comes out to be 3/112(2x+1)^4/3(8x-3)+C
Any help is appreciated!

 

[mmm4444bot]

… du=(2x+1)^1/3

That would be 2/3(2x+1)^2/3

because u=(2x+1)^1/3

That makes sense to me …

Not to me, heh.

What are you saying would be 2/3(2x+1)^2/3 specifically?

Why does u = du? Typos? :wink:

… x(2x+1)^1/3= 3/2 u³(u³/2-1/2)

The only part that does not make sense to me is how did they get the 3/2u³?

I'll list some steps (from the beginning), and I'll leave the final substitutions and simplification for you.

u = (2x + 1)^(1/3)

Solving that for x gives x = 1/2 * u^3 - 1/2

Taking the derivative of each side of that gives

du = 1/3 * (2x + 1)^(1/3 - 3/3) * (2) dx

which simplifies to

du = 2/3 * (2x + 1)^(-2/3) dx

Solving that for dx gives:

dx = 3/2 * (2x + 1)^(2/3) du

Using a property of exponents, note:

(2x + 1)^(2/3) = [(2x + 1)^1/3]^2 and that's u^2

Therefore:

dx = 3/2 * u^2 du

Now, carefully make the substitutions for x, (2x + 1)^(1/3), and dx -- in the original indefinite integral -- then simplify. :cool:

 

[mmm4444bot]

Here's another approach, for your consideration. :cool:

\(\displaystyle \displaystyle \int{x \cdot \sqrt[3]{2x + 1}} \; \text{dx}\)

\(\displaystyle u = 2x + 1\)

\(\displaystyle \text{du} = 2 \; \text{dx}\)

\(\displaystyle \displaystyle \frac{1}{4} \int{(u - 1) \; u^{1/3} \; \text{du}}\)

Expand the integrand.

\(\displaystyle \displaystyle \frac{1}{4} \int{u^{4/3} - u^{1/3}} \; \text{du}\)

Integrate term by term.

\(\displaystyle \displaystyle \frac{1}{4} \int{u^{4/3}} \; \text{du} \; - \; \frac{1}{4} \int{u^{1/3}} \; \text{du}\)

\(\displaystyle \displaystyle \frac{3}{28} \; u^{7/3} - \frac{3}{16} \; u^{4/3} + C\)

Back substitute for u.

\(\displaystyle \displaystyle \frac{3}{28} (2x + 1)^{7/3} - \frac{3}{16} (2x + 1)^{4/3} + C\)

Simplify.

\(\displaystyle \displaystyle \frac{3}{112} \; (2x + 1)^{4/3} \; (8x - 3) + C\)

 

[JeffM]

I have a problem like this : x(2x+1)^1/3
We need to find the antiderivative.
The step I don't understand, which is in the book, is this:
du=(2x+1)^1/3
That would be 2/3(2x+1)^2/3 because u=(2x+1)^1/3
That makes sense to me, as it is the derivative of (2x+1)^1/3.

First of all, you never say what concept you are finding difficult to understand. That makes it impossible to help you understand it. Telling us the section heading in your text might give us a clue what concept is being discussed.

I suspect that the concept has something to do with eliminating radicals.

\(\displaystyle du = (2x + 1)^{1/3} \ dx\).

Your apparent guess that that substitution comes from the derivative of

\(\displaystyle \dfrac{2}{3} * (2x + 1)^{2/3}\) makes no sense at all.

\(\displaystyle v = \dfrac{2}{3} * (2x + 1)^{2/3} \implies\)

\(\displaystyle \dfrac{dv}{dx} = \dfrac{2}{3} * \dfrac{2}{3} * 2 * (2x + 1)^{\{(2/3) - 1\}} = \dfrac{8}{9} * (2x + 1)^{-1/3} \ne (2x + 1)^{1/3}.\)

You need to check your book. I very strongly suspect that mmm is correct: your book actually starts by defining u rather than du, as in

\(\displaystyle u = (2x + 1)^{1/3}.\)

Why might they choose that? Because now there is a simple variable being exponentiated rather than an expression. Furthermore when we compute du we get

\(\displaystyle u = (2x + 1)^{1/3} \implies\)

\(\displaystyle \dfrac{du}{dx} = \dfrac{1/3} * (2x + 1)^{\{(1/3) - 1\}} * 2 = \dfrac{2}{3} * (2x + 1)^{-2/3} = \dfrac{2}{3} * u^{-2} \implies\)

\(\displaystyle du = \dfrac{2}{3} * u^{-2} \ dx \implies dx = \dfrac{3}{2} * u^2 \ du.\)

Again we have a variable being exponentiated rather than an expression. Of course, to get an integral in terms of u and du, we must also find x in terms of u.

Changing an integral in x and dx into a more convenient integral without radicals in u and du is the concept being discussed, but how can I be sure when you say nothing about what concept is being discussed.

 

[calc67x]

Fantastic help!

This is much simpler, what I was looking for!

Here's another approach, for your consideration. :cool:

\(\displaystyle \displaystyle \int{x \cdot \sqrt[3]{2x + 1}} \; \text{dx}\)

\(\displaystyle u = 2x + 1\)

\(\displaystyle \text{du} = 2 \; \text{dx}\)

\(\displaystyle \displaystyle \frac{1}{4} \int{(u - 1) \; u^{1/3} \; \text{du}}\)

Expand the integrand.

\(\displaystyle \displaystyle \frac{1}{4} \int{u^{4/3} - u^{1/3}} \; \text{du}\)

Integrate term by term.

\(\displaystyle \displaystyle \frac{1}{4} \int{u^{4/3}} \; \text{du} \; - \; \frac{1}{4} \int{u^{1/3}} \; \text{du}\)

\(\displaystyle \displaystyle \frac{3}{28} \; u^{7/3} - \frac{3}{16} \; u^{4/3} + C\)

Back substitute for u.

\(\displaystyle \displaystyle \frac{3}{28} (2x + 1)^{7/3} - \frac{3}{16} (2x + 1)^{4/3} + C\)

Simplify.

\(\displaystyle \displaystyle \frac{3}{112} \; (2x + 1)^{4/3} \; (8x - 3) + C\)

 

[Steven G]

I have a problem like this : x(2x+1)^1/3
We need to find the antiderivative.
The step I don't understand, which is in the book, is this:
du=(2x+1)^1/3
That would be 2/3(2x+1)^2/3 because u=(2x+1)^1/3
That makes sense to me, as it is the derivative of (2x+1)^1/3.
Then we have x=u³/2-1/2, which makes sense.
However, I then see where it says:
x(2x+1)^1/3= 3/2 u³(u³/2-1/2)The only part that does not make sense to me is how did they get the 3/2u³?
I would think that derivative of u³/2-1/2 would be 3/2u²
After this step, the rest of the problem is easily understood.
The answer comes out to be 3/112(2x+1)^4/3(8x-3)+C
Any help is appreciated!

If the problem was to find the antiderivative of x^1/3*(2x+1), this would be easy since you can multiply out and get 2x^4/3 + x^1/3 which is easy to integrate. So if you let u = (2x+1), then (2x+1)^1/3 becomes u^1/3 and x=(1/2)(u -1). Not that u-1 has no power which is what we wanted. du = 2dx. Now you can multiply everything out and get each term as a power of u which is easy to integrate.

This is a skill that you must have as it makes these problems easy to solve.