Partial Fractions: Simplifying Integrals with Higher Degree Numerators

[nameVoid]

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


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


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


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


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

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


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


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


$$4ln|x|+ln|x^2+4|+\frac{3}{2}arctan(x/2)+G$$
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[Mark44]

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.

 

[Bohrok]

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.

 

[nameVoid]

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

 

[Mark44]

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.