Welcome back.
I want to do a few more examples
of integration by parts.
So let's go ahead and get started.
Here we have our formula
for the integration by parts.
So we're going to be given an integral.
We'll let part of it equal u
and the remaining part equal dv.
Once we choose our u, we'll find differential u.
And once we choose dv, we'll integrate to find v.
Then we will apply the integration by parts formula
and see if the resulting integral
is easier to determine
or if maybe we have to use integration by parts again.
Again, I did review these on the last video,
so you may want to watch that video first.
Let's take a look at this example.
Here's our integration by parts formula.
Let's start by letting this x equal our u.
Therefore, our differential u or du
would equal the derivative of this
which would be one times dx.
Now the remaining part of the integral,
x plus two to the seventh dx
would have to be our dv.
Therefore, in order to find v,
we would have to integrate both sides of this equation.
When we integrate both sides of this to find v,
you may wonder if we have to perform substitution.
But since u would be x plus two
and u prime would be equal to one,
u substitution is not necessary.
Therefore, when we apply the power rule for integration,
we'll add one to the exponent
and then divide by that new exponent.
Now let's apply our integration by parts formula
and see if this integral is easier to integrate.
So this is equal to u times v.
So when I write this product, I'm going
to rewrite it as one-eighth x
times x plus two to the eighth power
minus the integral of v du.
Well, here's our v.
And du is equal to dx.
So in fact, this x was a problem
on this original integral.
And now the x is no longer there.
We can integrate this just as we integrated this to find v.
We may be thinking we have to perform integration
by substitution.
But if this is our u, du is going to be equal to one dx.
So in fact, we don't have to do the substitution.
So applying the power rule,
we'll add one to this exponent,
divide by the new exponent,
and add our constant of integration.
Let's go ahead and clean this up.
Here we'd have the denominator of 72
and then, of course, our plus C.
Let's take a look at this example.
Now the first thing I notice here
is we could apply the power property of logarithms
and rewrite this as the integral of two x
natural log x dx.
Now notice I chose to leave the two
inside the integral symbol.
I could factor it out.
But for me, it's easier to keep track
of all these substitutions if I leave it in this form.
So we need to pick our u.
Typically when the log function is involved,
that's going to be our u.
Let's try it.
Again, if it doesn't work, we'll just switch
and try something else.
So du would equal one over x dx.
And whatever's left must be our dv.
So in order to find v, we have to integrate both sides.
Add one to our exponent and then divide by two.
So we'd have x-squared.
Let's apply our formula.
This is equal to u times v
minus the integral of v du.
This looks good.
This will simplify nicely.
So it's simplified to one factor of x.
So now we can apply the basic power rule.
So now we can apply the power rule of integration.
Add one, divide by two for our antiderivative.
Now I'll take a look at a couple definite integrals.
Here, if we let u equal two x plus one,
du will be rather simple.
So let's try that.
So differential u will equal two dx.
So the remaining part will have to be our dv.
Since we know we have to integrate,
let's write this as e to the power
of negative x dx.
Again, if I move this up, the numerator would change
the sign of the exponent on base e.
So to integrate this, we would have a u
of negative x and a du of negative dx
or negative du is equal to dx.
So when we integrate both sides of this equation,
we're going to have v equal to negative e
to the power of negative x.
So what we did here is we performed substitution
in order to integrate both sides.
So when we apply integration by parts formula,
we'll have our u times v
minus the integral of v du.
Let's clean this up.
When we integrate this, we're going to have to perform
u substitution as we did over here.
So we will have an antiderivative
of negative e to the negative x power.
And this is a definite integral,
so we have to evaluate this at one and zero.
So when we sub in a one, we'd have this.
And these are actually like terms.
We'd have a negative five e to the negative one minus.
Over here, remember e to the zero would be one.
So we'd have negative one minus two, or negative three.
And the final result would be
approximately 1.1606.
Now we can check this on the graphing calculator
since it is a definite integral.
Let's go ahead and do that.
So if we hit math option nine,
type in the original integrand,
we can verify our work.
Let's take a look at one more.
Right away when I see natural log x,
I'm going to let that part equal u.
du will equal one over x dx.
So if that's u, then all of this will have to be dv.
We'll rewrite this as x to the power
of negative two dx.
Integrate both sides to obtain a v value
of add one and divide by the new exponent.
And apply the integration by parts formula:
u times v minus the integral of v du.
Simplify this, plus.
This should be x to the power of negative two dx.
Applying the power rule here,
we'd have x to the negative one divided by negative one
which would give us a minus one over x.
We can evaluate it at e and one.
Remember that natural log e is equal to one
and natural log one is equal to zero.
The final result here is approximately .2642.
Again, we can check this on the graphing calculator,
but I won't take the time for it now.
I hope these additional examples were helpful.
Again, thank you for watching.