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## Calculus 2

# Indefinite integral of 1/x

AP.CALC:

FUN‑6 (EU)

, FUN‑6.C (LO)

, FUN‑6.C.1 (EK)

, FUN‑6.C.2 (EK)

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.

## Video transcript

What I want to do
in this video is think about the
antiderivative of 1/x. Or another way of thinking about
it, another way of writing it , is the antiderivative of
x to the negative 1 power. And we already
know, if we somehow try to apply that
anti-power rule, that inverse power
rule over here, we would get something
that's not defined. We would get x to the 0 over
0, doesn't make any sense. And you might have been
saying, OK, well, I know what to do in this case. When we first learned
about derivatives, we know that the
derivative-- let me do this in yellow-- the
derivative with respect to x of the natural log
of x is equal to 1 over x. So why can't we just say
that the antiderivative of this right over here is equal
to the natural log of x plus c? And this isn't
necessarily wrong. The problem here is that
it's not broad enough. When I say it's
not broad enough, is that the domain over here,
for our original function that we're taking the
antiderivative of, is all real numbers
except for x equals 0. So over here, x
cannot be equal to 0. While the domain over here
is only positive numbers. So over here, x, so
for this expression, x has to be greater than 0. So it would be nice
if we could come up with an antiderivative that has
the same domain as the function that we're taking the
antiderivative of. So it would be nice if we could
find an antiderivative that is defined everywhere that
our original function is. So pretty much everywhere
except for x equaling 0. So how can we rearrange
this a little bit so that it could be defined
for negative values as well? Well, one one
possibility is to think about the natural log of
the absolute value of x. So I'll put little
question mark here, just because we
don't really know what the derivative of
this thing is going to be. And I'm not going to
rigorously prove it here, but I'll I will give you kind
of the conceptual understanding. So to understand it, let's
plot the natural log of x. And I had done
this ahead of time. So that right over
there is roughly what the graph of the
natural log of x looks like. So what would the natural
log of the absolute value of x is going to look like? Well, for positive x's, it's
going to look just like this. For positive x's you take
the absolute value of it, it's just the same thing as
taking that original value. So it's going to look just
like that for positive x's. But now this is also going to
be defined for negative x's. If you're taking the
absolute value of negative 1, that evaluates to just 1. So it's the natural log of 1, so
you're going to be right there. As you get closer and
closer and closer to 0 from the negative
side, you're just going to take the
absolute value. So it's essentially going
to be exactly this curve for the natural log of
x, but the left side of the natural log of
the absolute value of x is going to be its mirror image,
if you were to reflect around the y-axis. It's going to look
something like this. So what's nice
about this function is you see it's defined
everywhere, except for-- I'm trying to draw it as
symmetrically as possible-- except for x equals 0. So if you combine this pink
part and this part on the right, if you combine
both of these, you get y is equal to the natural
log of the absolute value of x. Now let's think
about its derivative. Well, we already know what the
derivative of the natural log of x is, and for
positive values of x. So let me write this down. For x is greater than 0,
we get the natural log of the absolute value of x is
equal to the natural log of x. Let me write this. Is equal to the
natural log of x. And we would also know,
since these two are equal for x is greater
than 0, the derivative of the natural log of
the absolute value of x is going to be equal to the
derivative of the natural log of x. Which is equal to 1/x
for x greater than 0. So let's plot that. I'll do that in green. It's equal to 1/x. So 1/x, we've seen it before. It looks something like this. So let me do my best
attempt to draw it. It has both vertical and
horizontal asymptotes. So it looks something like this. So this right over here is
1/x x is greater than 0. So this is 1/x when
x is greater than 0. So all it's saying here, and
you can see pretty clearly, is the slope right over here,
the slope of the tangent line is 1. And so you see
that when you look at the derivative, the
slope right over here, the derivative should
be equal to 1 here. When you get close
to 0, you have a very, very steep
positive slope here. And so you see you have a very
high value for its derivative. And then as you move away
from 0, it's still steep. But it becomes less
and less and less steep all the way until you get to 1. And then it keeps getting
less and less and less steep. But it never quite gets
to absolutely flat slope. And that's what you see
its derivative doing. Now what is the natural
log of absolute value of x doing right over here? When we are out here, our
slope is very close to 0. It's symmetric. The slope here is essentially
the negative of the slope here. I could do it maybe clearer,
showing it right here. Whatever the slope
is right over here, it's the exact
negative of whatever the slope is at a symmetric
point on the other side. So if on the other side, the
slope is right over here, over here it's going to
be the negative of that. So it's going to be
right over there. And then the slope it just
gets more and more and more negative. Right over here, the
slope is a positive 1. Over here it's going
to be a negative 1. So right over here our
slope is a negative 1. And then as we get
closer and closer to 0, it's just going to get more
and more and more negative. So the derivative
of the natural log of the absolute value of
x, for x is less than 0, looks something like this. And you see, and
once again, it's not a ultra rigorous
proof, but what you see is that the
derivative of the natural log of the absolute value of
x is equal to 1/x for all x's not equaling 0. So what you're seeing,
or hopefully you can visualize, that
the derivative-- let me write it this way--
of the natural log of the absolute value of x is
indeed equal to 1/x for all x does not equal 0. So this is a much more
satisfying antiderivative for 1/x. It has the exact
the same domain. So when we think about what
the antiderivative is for 1/x-- and I didn't do a kind
of a rigorous proof here, I didn't use the definition of
the derivative and all of that. But I kind of gave you a visual
understanding, hopefully, of it. We would say it's
the natural log of the absolute
value of x plus c. And now we have
an antiderivative that has the same
domain as that function that we're taking the
antiderivative of.