# Indefinite Integration by exchange of variables

1. Apr 26, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

This is only an example but I do not understand what they are doing...

$$\int (4x+1)^{3} + (4x+1)^{2}+(4x+1) dx$$

2. Relevant equations

$$\int f(u) du = F(g(x)) + C$$

3. The attempt at a solution

Let

$$u = 4x+1$$

then

$$du = 4 dx$$

and

$$dx = \frac {1}{4}du$$

how did they get that dx was 1/4? There are no steps to explain this, it just lists them as having that value.

So after substituting the problem should look like this:

$$\int (u^{3} + u^{2} + u) * \frac {1}{4} du$$

which is this:

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

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

So I suppose the only portion that I really don't understand is how they got the 1/4 dx value out of seemingly nothing....

2. Apr 26, 2010

### zachzach

$$du = 4 dx$$ solve this for $$dx$$

3. Apr 26, 2010

### Asphyxiated

I guess I just didn't really think about it all that hard... I got another question though, but Ill post it in a new thread

4. Apr 26, 2010

### System

5. Apr 26, 2010

### Legendre

There appears to be many notations and definitions of "dy" where y is a dummy variable. Some authors chose to define dy, while others leave it at "no meaning". I came across one interpretation that might help you...

du/dx = 4 = 4 dx/dx

So, (1/4) du/dx = dx/dx

By suppressing the denominator, we get,

(1/4) du = dx

Keeping in mind that the denominator is still there. That is, "dy" is shorthand for "dy/dx".

I like this interpretation because it does not appear to manipulate du/dx like a fraction.

Hope this helps!