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# 𝘶-substitution: rational function

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
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?
• Not if you include complex numbers
though I don't know if they would ever really apply
• where is the video that expains d/dx (ln abs(x)) = 1/x , or can you expain.Thanks
• Why did du/u become (1/u)du? Or is that a given rule?
• think about it like this: du/u is the same thing as (du * 1) / u. Multiplication can always be pulled out to the side. remember when you would put (2/5)x to be the same as 2x/5?
• what happened to the du at the end?
• 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.
• At around he drops the du, what happens to it?
• 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.)
• is it ( ln|u|+c ) or ( log|u|+c )?
• 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
• 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,
• is it always nessacary to put in the c at the end like sal does at ?
If your teacher gives you a problem, don't you know if theres a constant?
• Yes, it is necessary - and this is the reason.
Integration and differentiation are inverse operations.
If given the function x², or x²+10 or x²-222, when you take the derivative you get 2x for all three. When you integrate 2x you will get back x² but you have lost the constant that may or may not have been there, so we include the C, where C could be 0 or -222 etc.

Later you will learn about the indefinite integral as a way of finding the family of functions that are the solutions to a partial differential equation, and your appreciation of the C will deepen.
• 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?
• 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.
• how to integrate e^5Logx-e^4logx whole divided by e^3logx-e^2logx
• 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.