# Having some trouble understanding a u-substitution

1. Jul 14, 2013

### lukka

i have reached a chapter in my pure mathematics text that introduces u-substitution as a method for solving integrals where in general if f(x) = dy/dx, an integral with respect to x may be transformed into an integral with respect to a related variable u such that: ∫ f(x) dx = ∫ [f(x) dx/du]du = y.

A number of examples are carefully illustrated to demonstrate this technique. There is one problem in particular whereby the worked solution is as follows;

Find; ∫ x√(3x-1)dx

Solution;

let: u = √(3x-1), thus.. dx/du = 2u/3 and let: x=1/3(u^2 +1).

→ ∫ [x√(3x-1) dx/du]du = ∫ [1/3(u^2+1)u 2u/3]du
= ∫ [2/9(u^4) + 2/9(u^2)]du

The problem i have here is that when i separate the variables and integrate the [dx/du = 2u/3] i get x=1/3(u^2) + c, and not x=1/3(u^2 +1) as shown in the book. i understand the further simplification from here and i can see that setting the constant of integration to one (c=1) to give x=1/3(u^2 +1) is perhaps some kind of trick that allows one to get the second term in terms of the derivative of x, w.r.t u? but i'm unable to figure out how the intuition for this works out, after all could that constant not be any real number?

I have not seen this kind of substitution before. i only solved for and substituted for u and du so far. Can anyone explain to me what kind of u substitution this is?

I am new to the forum as an official member and would like to take opportunity to say thank you to all the members and contributors to this site, i have read many informative and interesting problems and discussions over the years and as a resource has been astonishingly useful to me. Thanks again! :)

Last edited: Jul 14, 2013
2. Jul 15, 2013

### CompuChip

There is no integration involved (yet) in that step. If you substitute u for √(3x - 1) then you also need to rewrite the x preceding it, so you need to solve your definition u = √(3x - 1) for x:
u = √(3x - 1)
u² = 3x - 1
u² + 1 = 3x
1/3(u² + 1) = x

Now substitute this in x√(3x - 1) everywhere, where of course √(3x - 1) just gives you u again.

3. Jul 15, 2013

### lukka

Thanks for pointing that out CompuChip! Your explanation is sound. i guess i was solving it as a differential when i really didn't need to, just was not obvious to me at first. Silly mistake really! What variation of u-substitution is being used here? i have not seen the use of [dx/du]du before, when carrying out u-substituions in the past i have been substituting for u and du only and simplifying, i just feel like i've missed something here. thanks again for sharing your insight

Last edited: Jul 15, 2013
4. Jul 15, 2013

### CompuChip

The dx/du is used to transform dx into du, because you need to integrate w.r.t the new variable as well.

If you treat the derivative as an ordinary fraction - which you really shouldn't! But it often works - you could write
$\frac{dx}{du} \, du = \frac{dx}{\not{du}} \not{du} = dx$
This leads to notation like "Let u = sin(x), then du = cos(x) dx" that you may encounter in less rigorous texts.

More formally, if you go back to the definition of integration, you can get it from the chain rule.

Can you be a bit more specific on how you have "substituted for du"? Because I'm sure that you have used exactly this procedure but you probably weren't aware of it. E.g. if u = 3x - 1 you would simply get du = dx up to a constant pre-factor which you can do without formally computing du/dx.

5. Jul 19, 2013

### lukka

Thanks for your explanation, i understand what you are saying here. I can see now how is arises naturally from applying the chain rule..

You're absolutely right, i was using this method the whole time. it was just not obvious at the time since i was only computing functions up to a constant pre-factor as you've kindly pointed out. This resolves the confusion i was having. Thanks for your help, it is greatly appreciated!