www.mathcentre.ac.uk/.../9.9%20Integration%20that%20leads%20to%20log%20functions.mp4

We already know.

When we differentiate
log X.

We end up with.

Is one over X?

We also know that if we've
got Y equals log of a

function of X.

And we differentiate it.

Then what we end up with is the
derivative of that function over

the function of X.

Now the point about integrating
is if we can recognize something

that's a differential, then we
can simply reverse the process.

So what we're going to be
looking for or looking at in

this case, is functions that
look like this that require

integration, so we can go back
from there to there.

So let's see if we can just
write that little bit down

again and then have a look
at some examples.

So no, that is why is
the log of a function of

X then divide by The X
is the derivative of that

function divided by the
function.

So therefore, if
we can recognize.

That form.

And we want to integrate it.
Then we can claim straight away

that this is the log of the
function of X plus. Of course a

constant of integration because
there are no limits here.

So we're going to be looking for
this. We're going to be looking

at what we've been given to
integrate and can we spot?

A derivative. Over the
function, or something

approaching a derivative.

So now we've got the result.
Let's look at some examples.

So we take the integral
of tan XDX.

Now it doesn't look much like
one of the examples. We've just

been talking about, but we know
that we can redefine Tan X sign

X over cause X.

And now when we look at the
derivative of cars is minus sign

so. The numerator is very nearly
the derivative of the

denominator, so let's make it
so. Let's put in minus sign X.

Now having putting the minus
sign, we've achieved what we

want. The numerator is the
derivative of the denominator,

the top is the derivative of the

bottom. But we need to put in
that balancing minus sign so

that we can retain the
equality of these two

expressions. Having done
that, we can now write this

down.

Minus and it's that minus sign.
The log of caused X plus a

constant of integration, see.

We're subtracting a log, which
means we're dividing by what's

within the log.

Function, so we're dividing
by cause what we do know is

with. With dividing by
cars, then that's one over

cars and that sank. So this
is log of sex X Plus C.

Now let's go on again and have a
look at another example.

Integral of X
over one plus

X squared DX.

Look at the bottom and
differentiate it. Its derivative

is 2X only got X on top,
that's no problem. Let's make it

2X on top by multiplying by two.

If we've multiplied by two,
we've got to divide by two,

and that means we want a
half of that result there.

So now this is balanced out
and it's the same as that.

What we've got on the top
now is very definitely the

derivative of what's on
the bottom, so again, we

can have a half the log of
one plus X squared plus C.

We can even have this with
look like very complicated

functions, so let's have a
look at one over X Times

the natural log of X.

Doesn't look like what we've got
does it? But let's remember that

the derivative of log X is one
over X. So if I write this a

little bit differently, one over
X divided by log X DX.

Then we can see that what's on
top is indeed the derivative of

what's on the bottom, and so,
again, this is the log of.

Log of X plus a constant of
integration. See, so even in

something like that we can find
what it is we're actually

looking for. Let's take one more

example. A little bit contrived,
but it does show you how you

need to work and look to see.

If what you've got on the.

Numerator is in fact the
derivative of the denominator.

So let's have a look at this.

Looks quite fearsome as it's
written, but let's just think

about what we would get if we
differentiate it X Sign X. I'll

just do that over here. Let's
say Y equals X sign X. Now this

is a product, it's a U times by
AV, so we know that if Y equals

UV when we do the

differentiation. Why by DX
is UDV by the X

plus VDU by DX?

So in this case you is
X&V is synex, so that's X

cause X Plus V which is
synex.

Times by du by DX, but you was X
or do you buy X is just one?

So if we look what we can see

here. Is that the numerator?

Is X cause X sign X, which is
the derivative of the

denominator X sign X, and so
again complicated though it

looks we've been able to spot
that the numerator is again

the derivative of the
denominator, and so we can say

straight away that the result
of this integral is the log of

the denominator X Sign X.

Sometimes you have to look very
closely and let's just remember

if we just look back at this
one. But sometimes you might

have to balance the function in
order to be able to make it

look like you want it to look.
But quite often it's fairly

clear that that's what you have
to do. So do remember, this is

a very typical standard form of
integration of very important

one, and one that occurs a
great deal when looking at

differential equations.