We are asked to evaluate the integral.
Looking at the integrated function,
notice how we have a rational function,
where the degree of the numerator is higher
than the degree of the denominator.
And therefore, the first step is to perform long division.
To begin, we need to determine what times four x squared
is equal to 12 x cubed, which is three x.
Or if we need to, we can divide 12 x cubed,
by four x squared,
which will give us three x,
which is the first term in the quotient.
We place the three x over the x term here.
Now we multiply three x and the divisor.
Three x times four x squared is 12 x cubed.
Three x times 25 is 75 x, that's right,
plus 75 x here under the x term.
And now we subtract.
Instead of subtracting though, it'll be easier to add
the opposite, so we can change this to addition,
as long as we change the sign here and here.
We're now adding, this sum is zero,
and we have four x squared, and then 73 x
plus negative 75 x is negative two x,
giving us minus two x.
And we still have the plus 21.
Now we need to determine what times four x squared
is equal to four x squared, which of course is one.
And therefore the next term in the quotient is one,
so we have plus one.
Now I multiply one and divisor,
or one times four x squared plus 25,
of course, is four x squared plus 25.
Now we subtract by adding the opposites.
So it changes to addition, changes to negative,
and change this to negative,
and now adding, this sum is zero, we have negative two x,
and then 21 plus negative 25 is negative four,
giving us minus four.
This is the remainder because this is degree one,
and we have a degree two divisor.
Which means we can write the original integral
as the integral of three x plus one,
and then plus the remainder of the quantity
negative two x minus four, all over the divisor
of four x squared plus 25.
Because the denominator here does not factor,
and we have a degree one numerator,
we do not perform partial fraction decomposition,
the next step will be to write this
as four separate integrals.
This is equal to the integral of three x,
d x, plus the integral of one d x,
and now let's make this rational expression
up into the difference of two rational expressions,
where because the first term in the numerator
is negative two x, and we have plus here,
let's write minus and then factor out the two.
Then we have the integral of x divided by
the quantity four x squared plus 25 d x.
And now because the second term in the numerator
is negative four, let's write that as minus,
let's factor out the four.
Then we have the integral of one,
divided by the quantity four x squared plus 25 d x.
You know, looking at the integrals,
we can integrate the first two integrals
using basic integration formulas.
To integrate here we'll perform u substitution.
And this integral here fits the integration formula
shown here below.
So let's begin integrating.
The integral three x with respect to x
is three times x squared divided by two.
Plus the integral of one with respect to x is x.
Minus two, let's show the u substitution here.
Using this integration formula here, u is equal to
four x squared plus 25.
Differential u is equal to eight x d x.
We have x d x, so let's divide both sides by eight,
giving us one eighth d u equals x d x.
So performing the u substitution, this d x
is equal to one eighth d u,
let's factor out the one eighth,
and then we have d u, our numerator is now one,
and our denominator of four x squared plus 25 is u.
Now for this last integral, we have minus four
times the integral.
This integrated function does fit the form
of one divided by the quantity a squared plus u squared,
where a squared is the constant 25,
and u squared is four x squared.
So if we needed to, we could rewrite this denominator
as the quantity five squared plus the square of two x,
where again, a is equal to five,
u is equal to two x,
and therefore, differential u is equal to two d x,
dividing both sides by two, we have one half d u,
equals d x.
So performing the substitutions,
again d x gives us one half d u,
so we factor out the one half,
then we have d u.
And the numerator stays one, and again,
four x squared plus 25 is the same as the square of five
plus the square of two x, which does give us a squared
plus u squared.
And now we can finish integrating.
So we have three halves x squared,
plus x,
here two times one eighth simplifies to one fourth,
giving us minus one fourth.
The integral of one over u d u is equal to
natural log absolute value of u plus c,
giving us natural log absolute value
of the quantity four x squared plus 25.
Then we have minus four times one half is two,
but then when integrating, we have one over a,
which is one over five, or one fifth.
Two times one fifth is two fifths.
And then we have arc tangent,
u divided by a is two x divided by five,
or two fifths x.
And then finally we have the plus c.
I hope you found this helpful.