How do you integrate the function $\frac 1{x^2 - a^2}$, where a is a constant?

Question

My problem with integrating by parts is that it always ends up being recursive, as I feel like I'm going in loops. Would appreciate it if someone can help me understand the process.

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

I know the answer is supposed to be $\dfrac 1{2a}\ln\left|\dfrac {x-a}{x+a}\right|+C$, but I can't figure out how to derive it.

mjh has given the best answer, but alternatively, notice that $x^{2}-a^{2} = (x+a)(x-a)$ and use partial fractions. — Matthew Cassell, Dec 03 '14 at 14:28

Travis Willse · Answer 1 · 2014-12-03T15:18:04.180

6

The integrand factors as $$\frac{1}{(x - a) (x + a)},$$ so we can decompose it via the Method of Partial Fractions--- $$\frac{1}{(x - a) (x + a)} = \frac{K}{x + a} + \frac{L}{x - a}$$ ---and cross-multiply and collect like terms to solve for $K$ and $L$. Then, we can integrate the terms separately using the elementary formula $$\int \frac{dx}{x + b} = \log|x + b| + C$$ (and then applying the usual identity for the sum of logarithms).

Solving gives $$K = - \frac{1}{2a} \qquad \text{and} \qquad L = \frac{1}{2a}.$$

Alternately, one can substitute $$x = a \cosh t,$$ which gives an especially nice integral, or, as mjh points out in the comments, $$x = a \sec \theta.$$

edited Dec 03 '14 at 15:18

answered Dec 03 '14 at 14:33

Travis Willse

99,363

I would've never thought to substitute acosht , btw, what is the "h" in there? Thanks for the suggestions! – user1746848 Dec 03 '14 at 14:50
$\cosh$ is the hyperbolic cosine function, defined to be $\cosh x := \frac{1}{2}(e^x + e^{-x})$, which you can think of as the "even part" of the exponential function $e^x$. And you're welcome, I hope you found them useful! – Travis Willse Dec 03 '14 at 14:57

score 5 · Accepted Answer · answered Dec 03 '14 at 14:30

5

You have to compute $A$ and $B$ such that $$\frac1{x^2-a^2}=\frac A{x-a}+\frac B{x+a}$$

To do that, write $$\frac A{x-a}+\frac B{x+a}=\frac{Ax+Aa+Bx-Ba}{x^2-a^2}$$

Then we have an equality of polynomials: $Ax+Bx+Aa-Ba=1$

We conclude that: $$A+B=0$$ $$Aa-Ba=1$$

Solve this system for $A$ and $B$.

answered Dec 03 '14 at 14:30

ajotatxe

65,084

I completely forgot about the partial fractions technique and use integration by parts instead, which didn't work out. Thanks for the clear explanation! – user1746848 Dec 03 '14 at 14:51

score 3 · Answer 3 · edited Dec 03 '14 at 14:59

3

Subtitute $x=\sqrt a\sec(u).$ Then, $dx=\sqrt a\sec(u)\tan(u)\ du.$ Then we have

$$\int\frac{\sec(u)\tan(u)}{a\tan^2(u)}du$$

$$=\frac{1}{a}\int\frac{\sec(u)}{tan(u)}du$$

You can probably do it from here :)

edited Dec 03 '14 at 14:59

homegrown

3,678

answered Dec 03 '14 at 14:35

Kurtbusch

632

How do you integrate the function $\frac 1{x^2 - a^2}$, where a is a constant?

3 Answers3