# Integral of a Fraction

How to find the integral of a fraction: 2 examples and their solutions.

## Example 1

### Example

### Solution

x^{2} - 3x + 5 cannot be factored.

So simplify (x^{2} - 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 (x^{2} - 3x + 5)/(x - 1) = x - 2 + 3/(x - 1).

Solve the integral.

The integral of x is [1/2]x^{2}.

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]x^{2} - 2x + 3 ln |x - 3| + C is the answer.

## Example 2

### Example

### Solution

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|.

Linear Change of Variable Rule

This integral is an indefinite integral.

So write +C.

So 2 ln |x| + ln |x - 1| + C is the answer.