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# 𝘶-substitution: rational function

Another example of using u-substitution. Created by Sal Khan.

## Want to join the conversation?

• At the very end ( ish), could the absolute value signs be dropped, since x^4+7 will always be positive?
(69 votes)
• Not if you include complex numbers
though I don't know if they would ever really apply
(4 votes)
• What exactly does dy/dx mean? I know it's the derivative of y with respect to x, but what do dy and dx mean individually, and what does it mean to multiply an expression by them individually? Around , Sal says du/dx is not a fraction, but that we can still manipulate it like one. Why? Sorry for all the questions at once, but this has been a great mystery to me ever since Calc I, and no one has ever given me an explanation,
(32 votes)
• So…yeah I have always been annoyed by this too, but apparently there is a long and difficult proof that shows that we can treat it like a fraction. If you really want to see it, I would suggest that you ask your professor, but meanwhile, I found a really good explanation as to why it is not a fraction, and what dy/dx really means : http://math.stackexchange.com/questions/409983/what-is-wrong-with-treating-dfrac-dydx-as-a-fraction
(22 votes)
• where is the video that expains d/dx (ln abs(x)) = 1/x , or can you expain.Thanks
(18 votes)
• what happened to the du at the end?
(15 votes)
• It does not become a constant. Recall estimating definite integrals by adding up areas of rectangles. It represents the infinitesimally small width of a rectangle as the number of rectangles becomes infinite. In indefinite integrals it just tells you what the variable is. In substitution, it relates derivatives of the variables x and u.
(5 votes)
• At around he drops the du, what happens to it?
(3 votes)
• du means "with respect to u". After you integrate with respect to u, you've dealt with it.

Consider where it comes from. If y is some function of x, when you take the derivative you replace y with dy/dx. The dx seems to come out of nowhere, so that's where it goes when you integrate. (It still means the same, as you take the derivative of y with respect to x.)
(5 votes)
• is it ( ln|u|+c ) or ( log|u|+c )?
(2 votes)
• Professional mathematicians assume that the base of a log is e, not 10. So, at this level of study, unlike earlier courses, you should assume all logs have a base of e unless otherwise specified.

Most professional mathematicians do NOT use "ln" but instead use "log" for the natural log. For the common log they use "lg" or "log₁₀".

Thus, the answer to your question is that `log |u| + C` and `ln|u| + C` mean the same thing, `logₑ |u| + C`
(8 votes)
• dy/dx as a quotient is very much intuitive , but then why do teachers always say that we must never think of it as a ratio we get when we divide dy by dx? If we are not to think of it like that then how can we multiply both sides and cancel out the dxs?
(2 votes)
• The early emphasis that dy/dx is not a fraction is for pedagogical reasons. Early in the study of calculus, it's important to grasp the concept that dy/dx is a limit, not a fraction, because this is one of the foundations upon which all of calculus is built. Later we learn that certain problems can be solved by treating dy and dx as distinct entities that can be manipulated in equations as if they were variables.
(3 votes)
• how to integrate e^5Logx-e^4logx whole divided by e^3logx-e^2logx
(2 votes)
• Let's cancel some of the e's first. We can factor this as e^2( e^3 logx-e^2 logx)/[e^2 (e logx-logx)]
The e^2 cancel and we're left with [e^3 logx-e^2 logx]/[e logx-logx].

A rule of logarithms turns this into [log(x^e^3)-log(x^e^2)]/[log(x^e)-logx].
Now this is a difference of logarithms, which turns this into log(x^(e^3-e^2))/log(x^(e-1))
Applying alog(x)=log(x^a) gives [(e^3-e^2)/(e-1)]•log(x)/log(x).
The log(x)'s cancel and we're left with just [e^3-e^2]/[e-1].
Factoring the numerator gives e^2[e-1]/[e-1].
The [e-1]'s cancel, and your whole expression is reduced to just e^2.
This is a constant, so its integral is xe^2 +C.
(3 votes)
• Are there any videos here explaining u-substitution instead of giving examples? This did not help me
(3 votes)
• 9 At , could we ignore the absolute value (or mod) because for any value of x the number inside the 'brackets' would be positive...........?
(3 votes)
• You can, but it is recommended you don't, as the problem is concerned with finding out the value of the integral, the general value of the integral of 1/u is log|u| + constant, it shows you know the integral, which is essential.
(1 vote)

## Video transcript

So we want to take the indefinite integral of 4x^3 over x^4 plus 7 dx. So how can we tackle this? It seems like a hairy integral. Now the key inside here is to realize you have this expression x^4 + 7 and you also have its derivative up here. The derivative of x^4 plus 7 is equal to 4x^3. Derivative of x^4 is 4x^3; derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. U-sub -- I'll just write u- -- I'll write the whole thing. U-Substitution could be the tool of choice. So given that, what would you want to set your u equal to? And I'll let you think about that 'cause it can figure out this part and the rest will just boil down to a fairly straightforward integral. Well, you want to set u be equal to the expression that you have its derivative laying around. So we could set u equal to x^4 plus 7. Now, what is du going to be equal to? du, I'm doing it in magenta. du, well it's just going to be the derivative of x^4 plus 7 with respect to x, so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- and I'll just rewrite it up here so that it becomes very obvious; our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? Well that's just going to be equal to the natural log of the absolute value -- and we use the absolute value so it'll be defined even for negative U's -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!