41st  Top calculus topics 
In integral calculus, the use of partial fractions is required to integrate the general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of partial fractions. Each partial fraction has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a function. If the denominator is a 1stdegree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nddegree polynomial or a power of such a polynomial, then the numerator is a 1stdegree polynomial.
Contents 
The substitution u = ax + b, du = a dx reduces the integral
to
The same substitution reduces such integrals as
to
Next we consider such integrals as
The quickest way to see that the denominator x^{2} − 8x + 25 is irreducible is to observe that its discriminant is negative. Alternatively, we can complete the square:
and observe that this sum of two squares can never be 0 while x is a real number.
In order to make use of the substitution
we would need to find x − 4 in the numerator. So we decompose the numerator x + 6 as (x − 4) + 10, and we write the integral as
The substitution handles the first summand, thus:
Note that the reason we can discard the absolute value sign is that, as we observed earlier, (x − 4)^{2} + 9 can never be negative.
Next we must treat the integral
First, complete the square, then do a bit more algebra:
Now the substitution
gives us
Putting it all together,
Next, consider
Just as above, we can split x + 6 into (x − 4) + 10, and treat the part containing x − 4 via the substitution
This leaves us with
As before, we first complete the square and then do a bit of algebraic massaging, to get
Then we can use a trigonometric substitution:
Then the integral becomes
By repeated applications of the halfangle formula
one can reduce this to an integral involving no higher powers of cos θ higher than the 1st power.
Then one faces the problem of expression sin(θ) and cos(θ) as functions of x. Recall that
and that tangent = opposite/adjacent. If the "opposite" side has length x − 4 and the "adjacent" side has length 3, then the Pythagorean theorem tells us that the hypotenuse has length √((x − 4)^{2} + 3^{2}) = √(x^{2} −8x + 25).
Therefore we have
and
