# Partial fractions

$$\int \frac{x^3+6x^2+3x+16}{x^3+4x} dx$$

$$\int \frac{x^3+6x^2+3x+16}{x(x^2+4)} dx$$

$$\frac{x^3+6x^2+3x+16}{x(x^2+4)}=\frac{A}{x}+\frac{Bx+C}{x^2+4}$$

$$x^3+6x^2+3x+16=A(x^2+4)+(Bx+C)x$$

$$x^3+6x^2+3x+16=Ax^2+4A+Bx^2+Cx$$
comparing coefficients..

$$A+B=6 , C=3 , A=4, B=2$$

$$\int \frac{4}{x}+\frac{2x+3}{x^2+4} dx$$

$$\int \frac{4}{x}+\frac{2x}{x^2+4} +\frac{3}{x^2+4}dx$$

$$4ln|x|+ln|x^2+4|+\frac{3}{2}arctan(x/2)+G$$

## Answers and Replies

Mark44
Mentor
For this to work for all values of x -
$$x^3+6x^2+3x+16=Ax^2+4A+Bx^2+Cx$$
there has to be an x3 term on the right side as well, which there isn't.

Before starting in with partial fractions decomposition, carry out the long division on your original integrand.

$$\frac{x^3+6x^2+3x+16}{x^3+4x} ~=~ 1 + \frac{some~quadratic}{x^3+4x}$$

So $$\int \frac{x^3+6x^2+3x+16}{x^3+4x} dx~=~ \int 1 + \frac{some~quadratic}{x^3+4x} dx$$
Now, do partial fractions decomposition.

Or instead of long division, rewrite the numerator so you have x3 + 4x in it so part of the numerator cancels with the denominator when you split it up.

thank you for explaining that mark thats a very usefull bit of info my text fails to mention

Mark44
Mentor
The thing to remember if you're going to use partial fractions when the degree of the numerator is >= degree of the denominator, carry out the division to get a numerator whose degree is < that of the denominator. In my reply, I show "some quadratic" in the numerator. That might or might not be correct. What is correct is that you'll get a polynomial of degree < 3.