Integral of a Fraction
How to find the integral of a fraction (rational function): 2 examples and their solutions.
Example∫ (x2 - 3x + 5)/(x - 1) dx
x2 - 3x + 5 cannot be factored.
So simplify (x2 - 3x + 5)/(x - 1)
by using synthetic division.
Write 1, -3, and 5.
Write the zero of (x - 1): 1.
↓: Write the left 1.
↗: 1⋅1 = 1
↓: -3 + 1 = -2
↗: -2⋅1 = -2
↓: 5 - 2 = 3
So (x2 - 3x + 5)/(x - 1) = x - 2 + 3/(x - 1).
Solve the integral.
The integral of x is [1/2]x2.
The integral of -2 is -2x.
Integral of a Polynomial
And the integral of +3/(x - 1) is
+3 ln |x - 1|.
Integral of 1/x
This is an indefinite integral.
So write +C.
So
[1/2]x2 - 2x + 3 ln |x - 3| + C
is the answer.
Example∫ [3x - 2]/[x(x - 1)] dx
See [3x - 2]/[x(x - 1)].
The denominator x(x - 1) is not a linear function.
And the order of the denominator, 2,
is higher than
the order of the numerator 3x - 2, 1.
Then change this fraction to partial fractions.
Set [3x - 2]/[x(x - 1)] = A/x + B/(x - 1).
The goal is to find A and B.
Combine A/x + B/(x - 1).
Then [(A + B)x - A]/[x(x - 1)].
Add and Subtract Rational Expressions
[3x - 2]/[x(x - 1)] = [(A + B)x - A]/[x(x - 1)]
See the numerators.
The x coefficients are always equal.
So A + B = 3.
The constants are always equal.
So -A = -2.
A + B = 3
-A = -2
Solve this system.
Then A = 2 and B = 1.
Substitution Method
[3x - 2]/[x(x - 1)] = A/x + B/(x - 1)
A = 2
B = 1
So [3x - 2]/[x(x - 1)] = 2/x + 1/(x - 1).
So (given) = ∫ (2/x + 1/(x - 1)) dx.
Solve the integral.
The integral of 2/x is
2 ln |x|.
The integral of +1/(x - 1) is
+ln |x - 1|.
Integral of f(ax + b)
This integral is an indefinite integral.
So write +C.
So
2 ln |x| + ln |x - 1| + C
is the answer.