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## AP®︎ Calculus BC (2017 edition)

### Unit 9: Lesson 4

Indefinite integrals of sin(x), cos(x), eˣ, and 1/x

# Indefinite integral of 1/x

In differential ​calculus we learned that the derivative of ln(x) is 1/x. Integration goes the other way: the integral (or antiderivative) of 1/x should be a function whose derivative is 1/x. As we just saw, this is ln(x). However, if x is negative then ln(x) is undefined! The solution is quite simple: the antiderivative of 1/x is ln(|x|). Created by Sal Khan.

## Want to join the conversation?

• What is the antiderivative of a constant to the power of negative one? Like 2^-1? Is it ln( l2l ) or something similar? •   A constant to the power of a constant, a constant divided or multiplied by a constant, a constant added to or subtracted from a constant, etc. is still just a constant.

If the value doesn't change, when you change x, you have a constant.

The graph of 2^-1 (which is 1/2 or 0.5) just like all constants is a straight horizontal line (It doesn't change with x).

The antiderivative of a straight horizontal line is a line with a slope.
i.e. integral(k dx)= k * x+C

Just to prove it works here:
remember d/dx(0.5x+c) = 0.5 = 2^-1

Hope this makes sense
• What is the difference between lnx and ln|x| ?
Thank you . •  Suppose `x` is a real number. We denote by `|x|` the absolute value of `x`. If `x ≥ 0`, we have `|x| = x`, and if `x < 0`, we have `|x| = -x`.

Denote the natural logarithm of a positive real number `x` by `log(x)`. The function defined by `log(|x|)` is the composition of the logarithm function with the absolute value function. In other words, it is the function defined by `log(|x|) = log(x)` if `x > 0`, and `log(|x|) = log(-x)` if `x < 0`.
• Why is ∫ 1/x dx not equal to, say, ln |2x|, or ln |3x|? For both of these, d/dx = 1/x, right? • At : Why do we search for ln |x|? at that moment? • • No. They are not the same.
Absolute value means the same thing the distance from 0.
Mod is short for modulo. The modulo operation means the remainder of a division.
Thus:
6 mod 3 = 0
7 mod 3 = 1
8 mod 3 = 2
9 mod 3 = 0

Whereas
| - 9 | = 9
and
| 2 + 3 𝑖 | = √13

NOTE: Your confusion is coming from the fact that the absolute value is also called the modulus. But that is not the same as the modulo (which is what mod stands for). Also note that the term modulus has other uses in mathematics.
• I've searched the antiderivative of x^-1, and it says it is log(x)+c not ln(x)+c. Why is this so?
(1 vote) • Professional mathematicians use the natural log, not the common log, as the assumed log. Most professional mathematicians do not use the notation "ln" for the natural log. Thus, at this level of study "log" without a base specified means the natural log.

Therefore, `log (x) + C` and `ln (x) + C` mean the same thing.

Note: it is very unusual to use any other base for a log in calculus than base e. There are a few areas of study where the binary (base 2) log is used, but other than those we nearly always we base e -- the math is just much easier with e as the base.
• If my x is negative for 1/x, can't I just factor out -1 and find the integral of 1/x, which is log(x) • AT Sal says the derivative of ln|x| is 1/x for all except o but we can clearly see that the derivative are varied, for x> 0 its 1/x and for x<0 its -(1/x). But how does Sal say its the same derivative ? • • At , why is the domain only positive numbers? Can't we take the log of negative numbers? • Yes, we can take the log of negative numbers, but it is a bit difficult and involves nonreal complex numbers. Thus, depending on what we are using the integral to represent, it often makes sense to place the absolute value around the argument as Sal did. But, that is not a strict requirement DEPENDING on what you are modeling with the integral or what your purposes might be.

log|x| is equivalent to saying Re{log (x)} (where Re represents the real portion of...)

However, if you have some application where the imaginary portion has meaning and significance, you don't have to put in the restriction.

For reference sake, here is the basics of the log of a negative number:
``Given that b > 0log (-b) = log(b) + iπ``

The base is, of course, e. For other bases:
``Given that b > 0 and k≠0 and k≠1logₖ(-b) = logₖ(b) + iπ logₖ(e)``