Hi everyone! Welcome back to integralcalc.com. We're gonna do a u-substitution problem this
morning. This was actually one of our questions submitted
by our viewers, Anna and the problem is the integral of four x over the square root
of x squared plus one and then dx of course. The way that I guessed this might be u-substitution
problem, first of all, I don't know just how to take the integral of this without modifying
it somehow. This is way too complicated to just take the integral straight out. I know
it's not partial fractions because you can't factor the denominator here. I am gonna try
to shy away from integration by parts because it's a lot harder to do integration by parts
when you have a fraction. It's much more complicated. So, my first guess is u-substitution because
you have something more complicated in the denominator here that I'm thinking we might
be able to plug u in for and then if you take the derivative of the bottom, you'll
get two x which might end up canceling with this four x so i'm thinking that we
might at least try using u-substitution. So let's just go ahead and try it.
I'm going to plug in, Im gonna assign u to x squared plus one and then (oh, cellphone. sorry.),
and then once I've assigned u to x squared plus one here, of course I need to take the
derivative of u which is gonna be du and the derivative of x squared plus one of
course, we bring the two out in front of the x here and we have two x. We subtract one from the
exponent which just leaves us with one which we don't have to write it so the derivative
is two x, the one cancels because the derivative of a constant is just zero so we have two x.
And then whenever we take du, we always have to remember to just tap dx on to the end
here because the whole reason we're taking this is so that we can solve this equation, du equals
two x dx for dx, so that we can plug it back in for dx up here. So the way that
we solve for dx of course is by dividing both sides by two x so we divide both sides
by two x everything cancels over here and we get dx equals du over two x.
So let's go ahead and plug it back into the integral here and see if we can actually end
up simplifying this. So we have the four x on top and remember, we assigned u to x squared
plus one so we have the square root of u, and then dx. Remember we solved and we got du over
two x so we plug du over two x back in for dx up here.
So now this is our new integral. Looks like we're gonna get lucky here and we're gonna
actually be able to cancel some things so you can see we have an x in the numerator
and an x in the denominator and those are both gonna cancel. The four and the two, we've
got four over two which will simplify to two, so let's go ahead and rewrite our integral
up here. It's gonna be two, remember because we've got four over two and then we have one
over the, so two times, over the square root of u du is our simplified integral.
Now we want to still simplify this a little bit further before we take the integral. The
first thing we're gonna do, we're gonna bring the two out in front here, because everything
in this integral is multiplied together, we have two times one over the square root of u times du.
Since everything here is multiplied together, this actually is one term. It's all connected.
And when that happens and you have a coefficient, like this two in front of everything, you
can move it out in front of the integral so that we don't have to deal with it inside
of the integral and we can simplify. So we're left with one over the square root of u du.
And now the way that I like to simplify this, whenever possible, I like to convert fractions
to non-fractions because I think it's a lot easier to take the integral of something that's
not a fraction. So I'm gonna go ahead and try to do that here. The way that we're gonna
do that, we know this is a rule up here, the square root of x is the same as x to the
one half. These are ident… These are equivalent or equal so we have two times the integral
of one over u to the one half and then of course du.
So now that we have u to the one half, we can move this to the numerator of this fraction
and the way that we do that, our second rule here, and I do this kind of conversion a lot
where I change the square root to one half and then move it to the top to make it easier
on myself. One over x let’s say x squared. One over x squared is the same thing as x to
the negative two. I can move this to the numerator, by changing the sign on the exponent from
a positive two to a negative two. So I'm gonna move this to the top and just make this a
negative one-half. So I end up with two times the integral of u to the negative one-half du.
Okay so now I'm finally feeling comfortable. This is something that I feel like I can take
the integral of. Let's go ahead and erase everything over here. Now we're gonna go ahead
and take the integral. Write your two out in front here and then drop a big parenthesis.
If you ever bring a coefficient or anything else outside an integral like this, you just
have to write it out and then big parenthesis to multiply it by everything that ends up
being inside the integral here when we take the integral. So two times, and this is
now simpler. Whenever we take the integral, of course we add one to the exponent first
so negative one-half plus one is positive one-half and then we always do the coefficient
on the term here which is one implied, right? This is actually one u to the negative one
half. We just don't write it because it's implied. So we have one, which is the implied
coefficient, divided by the new exponent which is one-half, so one over one-half. And then
of course we do, plus c. We always add that to the end whenever we're taking the integral
to account for any constant that may have been there.
So now we can simplify this. We have two times one divided by one-half is two and then u to
the one half plus c. So two times two here, we can say four. Remember our rule up here, u to
the one-half is the same thing as the square root of u so we have four times the square
root of u plus c. And now what we need to do is plug in x squared
plus one back in for u so four times the square root of x squared plus one plus c.
And that's actually our final answer. Just for style points though, if anyone is
interested the way that we know that that ends up being true if you ever are unsure
and you wanna check yourself, let's go ahead and just double-check here. So our final answer
was four times the square root of x squared plus one plus c. So now we want to double
check that our answer is right and of course we, (this is the anti-derivative of this function
here) so if we take the derivative of this function, we should get this back out. And
here's how we're gonna do that. So to take the derivative of this here, I'm gonna go
ahead and convert this back to, remember we said our rule, the square root of anything
is the same as that thing to the one half power. So we can go ahead and convert this
back to x squared plus one to the one-half, we can say plus c here.
So now to take the derivative, we would multiply this one half by the four out in front and
we get one half times four is two and then we'd say x squared plus one and we would
subtract one from the exponent, one half minus one is negative one half but then of course
we have to apply chain rule because this is more complicated than just x so we have
to multiply this whole thing by the derivative of this inside portion here. And the derivative
of this is just two x so we multiply by two x. And when we're taking the derivative, c is
just representing the constant like two or four, just a number and of course the derivative
of a number like that is just zero, so the c drops off then we could go ahead and take the derivative
again. And now we can just simplify this, right?
We have two times two here, everything here is multiplied together, so we've got four.
We can also bring this x out in front and then we have x squared plus one to the negative
one half. So remember the rule that we said before,
we have a negative exponent here, this term is in the numerator, if we move it to the
denominator, we can change this to a positive. You can go from the denominator to the numerator,
from the numerator to the denominator and just change the sign of the exponent. So we'll
end up with (I think I'd still have room down here), we'll end up with four x over x squared
plus one to the one half, we just changed this from a negative sign of the exponent
to a positive sign and now of course we know that x squared plus one to the one half
is the same thing as x squared plus one or the square root of x squared plus one,
so we see that our final answer is four x over the square root of x squared plus one which
is the, our original function here. So we have proven by just reversing this whole
process that this is the correct final answer. Just thought I could throw that in there for
you, guys. Have a great day!