Question

Integration problem. Spot my mistake.

Answer 1

I'm missing a "u" in the numerator.

Answer 2

I dunno, your handwriting is confusing me with those silly x's...

Lemme try to just work through this one and see if you can spot the mistake maybe.

Answer 3

So we want to show that:\[\huge \int\limits \frac{u^{-1/3}}{1+u} du \quad = \quad \int\limits \frac{3u}{1+u^3}du\]

Answer 4

Actually I'm going to change the variable on the right so we can make a substitution and match them up that way,\[\huge \int\limits\limits \frac{u^{-1/3}}{1+u} du \quad = \quad \int\limits\limits \frac{3x}{1+x^3}dx\]

Answer 5

Let's try to change the one of the left,\[\large \color{orangered}{u=x^3}, \qquad du=3x^2 dx\]Hmmm, let's take our u and solve for X.\[\large u=x^3 \qquad \rightarrow \qquad u^{1/3}=x \qquad \rightarrow \qquad \color{cornflowerblue}{u^{-1/3}=\frac{1}{x}}\]Andddd from our du equation let's do somethign clever, divide both sides by x.\[\large du=3x^2dx \qquad \rightarrow \qquad \frac{1}{x}du=3x dx\]

Now we'll replace our 1/x with the blue piece.\[\large \color{orangered}{u^{-1/3}du=3x dx}\]

Answer 6

From here, we should be able to plug in the two ORANGE pieces to make our replacement.

Answer 7

Kind of a weird question :D You have to mess around with a bunch of little stuff while you're making your substitutions.

Answer 8

Give me a sec.

Answer 9

I was getting this:

dx = 3du^2

Then i substituted into the original equation:
\[\Large \int\limits \frac{u^{-1}}{1+u^3 }\times 3du^2\]What do you think?

Answer 10

Ugh I think I just realized... you wrote the LEFT integral in terms of X didn't you?
I couldn't tell the difference between your x's and u's.
So all of my variables are the opposite of yours.
That's a little confusing :( my bad.

So we actually want to show this?\[\large \int\limits\frac{x^{-1/3}}{1+x}dx \quad = \quad \int\limits\frac{3u}{1+u^3}du\]
I dunno what du^2 is suppose to be. Hmmm let's see if we can find what went wrong.

So it looks like you started with:\[\large x=u^3\]
Ok that's good.

Answer 11

Taking the derivative WRT u gives us,\[\large \frac{dx}{du}=3u^2\]There is a process that allows us to write the du on the other side.
For our purposes let's simply think of it as multiplication.
So multiplying both sides by du gives us,\[\large dx=3u^2 du\]

Answer 12

It looks like you took the derivative WRT x maybe?

Getting confused by these weird differentials maybe? :)

Answer 13

Oh I think we still end up ok if we differentiate with respect to X. Lemme check real quick.\[\large x=u^3 \qquad \rightarrow \qquad 1=3u^2 \frac{du}{dx}\]Multiplying both sides by dx gives us,\[\large dx=3u^2 du\]

Answer 14

Oh yah I guess that works just fine.

Answer 15

OHH!!! RIHGTT!!!!!
I was doing it WRT to x. =_=
So we have to do WRT to u???

Answer 16

No look at my second to last post c: it looks like it still works WRT to x.

You just have to remember how the chain rule works c: How you'll have a du/dx term pop out.