Hi, everyone! Welcome back to integralcalc.com. Today we’re going to be talking about finding
integrals of inverse hyperbolic functions. And in this particular problem, we have the
integral of e^x over the square root of e^(2x + 1). And this becomes a hyperbolic function
because we’re going to end up making a u-substitution with this problem which is going to get us
to a point where there’s going to be a hyperbolic function involved in the integral.
So in this particular problem, we’re going to go ahead and make a u-substitution where
u = e^x. And then of course when we take the derivative to get du, we’ll get the same
thing, e to the x because the derivative of e^x is itself. So we get e^x dx. And then
we want to solve this for dx by dividing both sides by e^x. So we’ll get dx = du/e^x.
and now we can go ahead and plug these things back into our integral. So we know that e
to the x is equal to u so we plug that in to the numerator. And then in the denominator,
we have e^(2x + 1). Well, e^2x is the same thing as saying (e^x)(e^x) or that’s also
the same thing as (e^x)^2. So we can go ahead and change e^2x into u^2. So we get u^2 +
1 under the square root sign in the denominator. And then for dx, we’ll go ahead and substitute
and get du/e^x. Now, the goal of u-substitution is obviously to get everything in terms of
u but we still have e to the x hanging out here. What we need to remember is that we
set u = e^x in the first place. So even though it was left here in our equation with dx,
we can go ahead and change this e to the x into u because u = e^x. So we change it into
u and now we have a u in the denominator and a u in the numerator which will cancel with
one another. So what we’re left with is the integral of 1 over the square root of
u^2 + 1 du. And what’s great about this is that we’ve now got into a point, notice
that our integral currently matches the formula that we have up here and this formula tells
us that the integral of 1 over the square root of u^2 + 1 is going to be equal to the
inverse hyperbolic sign function of u + c. So we could go ahead and integrate and what
we would get is the inverse hyperbolic sign function of u and then we always have to add
c to account for that constant of integration. now, we need to go ahead and plug back in
for u. we know that it’s equal to e to the x so we’ll get the inverse hyperbolic sign
function of e^x and go ahead and leave that plus c there. But that’s it. That’s our
final answer. So the trickiest part about integrals that
involve inverse hyperbolic functions is converting integrals to a point where they match a formula
that gives you an inverse hyperbolic sign or an inverse hyperbolic function. So getting
your integral to this point here is going to allow you to easily take the integral because
we know that the integral of that is the inverse hyperbolic sine function of this new variable
here. So this goes for the other inverse hyperbolic functions as well. They all have a corresponding
formula that when you take the integral, it will give you the inverse hyperbolic cosine
function as well as of the other trigonometric identities. So we transformed our integral
to the point where we got this right here and we knew from the formula that the integral
of that was the inverse hyperbolic sine function so we could easily go ahead and take it. So
this is our final answer here. I hope this video helped you guys and I will see you in
the next one. Bye!