Indefinite Integration by exchange of variables

[Asphyxiated]

Homework Statement

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

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

Homework Equations

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

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...

 

[zachzach]

Homework Statement

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

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

Homework Equations

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

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...

du = 4 dx solve this for dx

 

[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

 

[System]

Your solution is correct.

 

[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!