**Integration Using Long Division** works best for rational expressions where the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator.

## Integration Using Long Division: Examples

**Example question #1**: Solve the following integral:

Step 1: Check to see if the degree in the numerator is greater than or equal to the degree in the denominator. For this example, it is, so we can continue.

Step 2: Divide the numerator by the denominator, using algebra:

Step 3: Rewrite the integral in the question with the new expression you got in Step 2:

Step 4: Solve the integral using the usual rules of integration:

- Apply the sum rule:
- Apply the common integral ∫
^{1}⁄_{x}dx = ln |x| + c: = 2x + 3 ln |x – 2| - Add a constant: 2x + 3 ln |x – 2| + C

Note: The sum rule states that the integral of the sum of two functions is the sum of their separate integrals:

∫[f(x) + g(x)] dx = ∫f(x)dx + ∫g(x) dx

**Example question #2**: Solve the following integral:

Step 1: Check to see if the degree in the numerator is greater than or equal to the degree in the denominator. For this example, it is, so we can continue.

Step 2: Divide the numerator by the denominator, using algebra:

We can’t actually divide this until we factor the denominator first:

Now we can perform the long division, dividing the numerator by the denominator:

The x – 1 is the quotient and 3x -2 is the remainder. Rewrite the solution above as “quotient + remainder/original factored denominator”:

Step 3: Solve the integral using the usual rules of integration: In addition to the sum rule and common integral ∫^{1}⁄_{x} dx = ln |x| + c: = 2x + 3 ln |x – 2|, we also need to apply the power rule to the “x” in front. The solution is: