how does one integrate: $\int \frac{1}{x^2-a^2}dx$

Question

how does one integrate: $$\int \frac{1}{x^2-a^2}dx$$

I know it looks very similar to the known formula $$\int \frac{1}{x^2+a^2}dx$$ but it doesn't help really.

note: Im not allowed to use the partial fraction method in the solution here.

Answer 1

If you substitute $x=a\tanh y$ you get $$\int \frac{1}{a^2\tanh^2y-a^2}a\text{sech}^2 ydy$$

This is similar to that $\tan$ substitution you might use for your known formula

Answer 2

Hint: $$\frac{1}{x^2-a^2} = \frac{1}{(x-a)(x + a)} = \frac{A}{x-a} + \frac{B}{x+a}$$

Answer 3

You can use :

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

Which is known, then $$\int \frac{1}{x^2 - a^2}dx = \frac{-\text{arctanh}(\frac{x}{a})} {a}$$

Answer 4

Note that

$${a\over x^2-a^2}={x+a-x\over x^2-a^2}={1\over x-a}-{x\over x^2-a^2}$$

Therefore

$$\begin{align} \int{1\over x^2-a^2}dx&={1\over a}\int{a\over x^2-a^2}dx\\ \\ &={1\over a}\int{1\over x-a}dx-{1\over a}\int{x\over x^2-a^2}dx\\ \\ &={1\over a}\ln|x-a|-{1\over2a}\ln|x^2-a^2|+C \end{align}$$

Remark: This looks a bit (or a lot) like partial fractions, but it's not the standard partial fraction method.

Answer 5

Well from the Pythagorean Identities of trig we have:

$\tan^{2}(\theta) + 1 = \sec^{2}(\theta)$

this tells us that if we let $x=a\sec\theta$ we can then reduce this to solving

$$\int\frac{a\sec\theta\tan\theta}{a^2\sec^2\theta-a^2}d\theta$$

I'm assuming you can take it from here.