integrate $\frac{x^2 + 81}{x^2 - 81}$

Question

I need help integrating $\frac{x^2 + 81}{x^2 - 81}$. I am assuming I need to use one of these tables: Integral Tables

I went ahead and factored out the denominator to $(x+9)(x-9)$ but what do I need to do to the numerator to get it into a format that will let me use the tables?

Answer 1

First, you need to simplify the function $\frac{x^2+81}{x^2-81}$. You can do this as $$\frac{x^2+81}{x^2-81} = \frac{x^2-81+81+81}{x^2-81} = 1+ \frac{81*2}{x^2-81} = 1 - \frac{9}{x+9} + \frac{9}{x-9}.$$ Then you can refer the Integral table for the details of the integral $\int \frac{1}{x} \text{d}x$ if you are dealing with the indefinite integral. If you are solving the definite integral, you should care about the bound of domain.

Answer 2

Hint: $$\frac{x^2+81}{x^2-81}=1+\frac{18\cdot 9}{(x-9)(x+9)} = \color{red}{1+\frac{9}{x-9}-\frac{9}{x+9}}. $$

Answer 3

$$=\frac{x^2-81+81+81}{x^2-81}=1+\frac{162}{x^2-81}=1+\frac{9}{x-9}-\frac{9}{x+9}$$