Question regarding partial fraction decomposition for $\int \frac{1}{x^2-4} dx$

Question

We want to integrate

$$\int \frac{1}{x^2-4} dx$$

I have tried the following:

$$\frac{1}{(x-2)(x+2)} = \frac{c_1}{x-2} + \frac{c_2}{x+2}$$

I wanted to find out the value of $c_1$ by multiplying with $(x-2)$:

$$\frac{(x-2)}{(x+2)} = c_1 + \frac{c_2 \cdot (x-2)}{x+2}$$

If I understood it correctly, we want $(x-2) = 0$ to cancel $c_2$ out.

$x \to 2$ leads to

$$\frac{2-2}{2+2} = c_1+ \frac{c_2 \cdot(2-2)}{2+2} \\ 0 = c_1$$

But an online integration calculator says $c_1 = \frac{1}{4}$ and $c_2 = - \frac{1}{4}$

Where did I go wrong?

$$$$

Answer 1

You went wrong when you wrote $\frac{(x-2)}{(x+2)}$ , the $(x-2)$ already cancels with the previous $(x-2)$

It should be instead $\frac{1}{(x+2)} = c_1+ \ldots$

Answer 2

Consider the partial fractions decomposition $\frac{1}{x^2-4} = \frac{1}{4}(\frac{1}{x-2}-\frac{1}{x+2})$. Then integrate as natural log functions.

$\int\frac{1}{x^2-4} = \frac{1}{4}\int(\frac{1}{x-2}-\frac{1}{x+2})$ = $\frac{1}{4}(\ln(x-2)-\ln(x+2)) = \frac{1}{4}\ln(\frac{x-2}{x+2})$

Answer 3

For an easier way to do the decomposition, note that you can express the numerator as a difference whenever:

1. The denominator contains terms which are separated only by a constant, as in your question: In general, for $$ \int\frac{1}{(x-a)(x-b)}dx,$$ you can write $$1=\frac{(x-a)-(x-b)}{b-a}$$ so that the integral is $$\frac{1}{b-a}\int\frac{(x-a)-(x-b)}{(x-a)(x-b)}dx$$$$=\frac{1}{b-a}\int \left(\frac{1}{x-b}-\frac{1}{x-a}\right)dx$$

In your case it is $$\int\frac{dx}{(x+2)(x-2)}$$$$=\frac14\int\frac{(x+2)-(x-2)}{(x+2)(x-2)}dx$$$$=\frac14\int\left(\frac{1}{x-2}-\frac{1}{x+2}\right)dx$$$$=\frac14(\ln|x-2|-\ln|x+2|)+C.$$

2. The denominator contains terms of the form $(P(x))(k\cdot P(x)+M)$ where $k,M$ are constants and $P(x)$ is a polynomial (preferably linear or quadratic). You then write $$\int\frac{dx}{P(x)(k\cdot P(x)+M)}$$ as $$\frac 1M\int\frac{(k\cdot P(x)+M)-k\cdot P(x)}{P(x)(k\cdot P(x)+M)}dx$$$$=\frac1M\int\left(\frac{1}{P(x)}-\frac{k}{k\cdot P(x)+M}\right)dx$$ and proceed further.