1. ## [SOLVED] integral evaluation

use partial fractions to derive the integration formula
$\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln\frac{a+x}{a-x}+c$

2. Originally Posted by vinson24
use partial fractions to derive the integration formula
$\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln\frac{a+x}{a-x}+c$
$\frac{1}{a^2 - x^2} = \frac{1}{(a - x)(a + x)} = \frac{A}{a - x} + \frac{B}{a + x}$.

Cross multiplying gives

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

So $A(a + x) + B(a - x) = 1$

$Aa + Ax + Ba - Bx = 1$

$(A - B)x + (A + B)a = 0x + 1$.

Equating like powers of x gives

$A - B = 0$ and $(A + B)a = 1$.

Solving these simultaneously gives $A = B$ and therefore $(A + B) = \frac{1}{a}$

$2A = \frac{1}{a}$

$A = \frac{1}{2a}$.

Therefore $A = B = \frac{1}{2a}$.

So the integral becomes

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

$= -\frac{1}{2a}\ln{(a - x)} + \frac{1}{2a}\ln{(a + x)} + C$

$= \frac{1}{2a}[\ln{(a + x)} - \ln{(a - x)}] + C$

$= \frac{1}{2a}\ln{\left(\frac{a + x}{a - x}\right)} + C$.