U-Substitution using Log Rule Video Example
Hello everyone I’m Tom from every step calculus dot com.
Im gonna do a U-substitution problem
that requires the log rule. And uh,
let’s get into it. index8() is my,
you have to put that in the end line here to get to my menu.
I’m already at U-substitution.
You scroll down with the cursor to get there.
And wait for it to load here for a second.
And we’re gonna enter our function. You have to press alpha before you enter anything in my
entry lines in my programs.
Alpha 8 times
X divided by
X squared plus 1
I will show you what you’ve entered. The reason you need to do a log rule is
notice in the denominator your have x squared plus one.
This is an exponent of 1. If you we’re to
transfer this up to the numerator the exponent of
one would become a -1. And in integration you always add,
the first thing you do is always add to the numerator.
And one plus minus 1 is zero so the answer will always come out to be
-1. That’s the reason anytime you see this without, without an exponent other than
one. The one is of course hidden. Um,
you know it’s a log problem.
So we’re going to press there.
Say its okay. Here’s the original
function. And all these you have to rewrite them. Notice there is a constant error
of 8. You have to bring that outside the integral, that’s very important. And we’re going to also
take the x
here and put it over by the dx.
And we’re going to put one in the numerator,
for X squared plus 1 in the denominator.
So U is x squared plus 1 the derivative of that is 2x.
Another trick to all these, took me a long time to figure out.
Was d you need to have the divisor of 2
You have to move that over as a divisor here, and leaving
this here. You’ll notice that X DX is the same as x dx.
That means it’s a U-substitution problem if it wasn’t it’s not a
U-substitution problem. I ask you that though just
to make sure you know what you’re doing. I say yes.
So we have i8 outside the integral and one over you
du number 2. Now we have a constant here at one half.
So we have to bring that outside the integral, which we do over here.
8 and 1/2 that equals 4.
And then we have four times log of U. Anything over one of U is
log of u, plus c.
And the answer is, after we substitute the u back in.
4 times log of x squared
plus 1 plus c. Pretty neat huh?
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