In this lesson, we'll be using
the method of U-substitution to evaluate
the indefinite integrals.
Analyzing the first integral, we need to determine
what to let u equal.
Notice how the numerator is a degree one polynomial,
and the radicand in the denominator is a degree two.
Since the derivative of a degree two polynomial
is equal to a degree one polynomial,
we will let u equal the radicand of 10x plus x squared.
In order to set u is equal to 10x plus x squared,
the denominator is now the square root of u,
which can be written as u to the power of 1/2.
Also notice we're left with the quantity
of x plus 5 times dx.
So the next step, we determine differential u,
what is equal to the derivative of 10x plus x squared
with respect to x times dx?
Which is 10 plus 2x, and then we have times dx.
So it may appear as if things are not going to match up
like we want, but notice 10 plus 2x factors,
because there's a common factor of 2.
Let's go ahead and factor out the 2,
which gives us differential u is equal to 2 times,
let's go ahead and change the order.
To factor out 2 from x, we have x,
if we factor out 2 from 10, we have 5,
giving us the quantity x plus 5,
and we still have times dx.
From here, because the integral has the quantity
x plus 5 times dx, we will divide both sides by 2
for a perfect match.
So if we divide both sides by 2,
we now know that 1/2 du
is equal to the quantity x plus 5 times dx.
Which means we can replace x plus 5 dx with 1/2 du.
So now we write the integral only in terms of u.
The quantity, x plus 5 dx is equal to 1/2 du,
let's factor out the 1/2.
Then we have du.
And then we have one over the square root of u,
again, because u is equal to 10x plus x squared.
Let's rewrite this as 1/2 times the integral
of one divided u to the power of 1/2 du.
In order to apply the power of integration,
we now need to write one divided by u
to the power of 1/2, as u to the power of negative 1/2,
which gives us 1/2 times the integral
of u to the power of negative 1/2 du.
And now we integrate with respect to u,
which gives us 1/2 times u to the power
of negative 1/2 plus 1 is positive 1/2
divided by positive 1/2 plus c.
Simplifying further, dividing by 1/2
is equivalent to multiplying by the reciprocal
of 2 over 1 which gives us 1/2 times 2 over 1,
times u to the power of 1/2 plus c.
Simplifying, we just have u to the 1/2 plus c,
which in terms of x, is the quantity 10x plus x squared
raised to the power of 1/2 plus c.
If we want, we can write this as the square root
of the quantity 10x plus x squared plus c.
We often use big F of x for the antiderivative,
or indefinite integral.
We can say big F of x is equal to the square root
of the quantity 10x plus x squared plus c.
This is the family of functions whose derivative
is equal to the given integrand function.
Looking at the next example,
analyzing the integrand function,
we have the degree two monomial outside the cube root,
and we have a degree three polynomial under the cube root.
And since a derivative of a degree three polynomial
is a degree two, we will let u equal
the radicand of 4x cubed plus 9.
Which means now this is the cube root of u,
which is equivalent to 2u to the power of 1/3.
And now we determine differential u
which is equal to the derivative of 4x cubed plus 9
with respect to x times dx,
which gives us 12x squared times dx.
Notice we have 3x squared dx, not 12x squared dx.
So we have an option here.
We could divide both sides by 4,
so that we have 3x squared dx on the right,
or we can just solve for x squared dx.
I'm going to go ahead and just solve for x squared dx,
because it's not always easy to determine
what to multiply or divide by for a perfect match.
Simplifying now, notice how we have
1/12 du equals x squared dx.
So going back to the integral,
we can factor out the 3 and now replace
x squared dx with 1/12 du.
So we're at the integral in terms of u.
Again, we factor out the 3,
then we have the integral of x squared dx is 1/12 du,
and then we have the cube root of u,
which is equal to 3/12 is 1/4
times the integral of the cube root of u
is equal to u to the 1/3 du.
And now we integrate with respect to u,
which is 1/4 times u to the power of 1/3 plus 1,
which is 4/3 divided by 4/3 plus c.
Dividing by 4/3 is equivalent to multiplying
by the reciprocal of 3/4, which gives us 1/4
times 3/4 times u to the power of 4/3,
which in terms of x is the quantity
4x cubed plus 9 raised to the power of 4/3 plus c.
Simplifying one more time,
we have 3/16 times the quantity 4x cubed plus 9
raised to the power of 4/3 plus c.
This can also be identified as big F of x,
which once again, is the family of functions
whose derivative is equal to the given integrand function.
I hope you found this helpful.