Question

Prove that `int 1/(x^2 - a^2) dx = 1/(2a) log |(x - a)/(x + a)| + c`.

Hence evaluate `int 1/(x^2 - 3) dx`.

Answer 1

`int 1/(x^2 - a^2) dx = int 1/((x - a)(x + a)) dx`

= `1/(2a) int ((x + a)-(x - a))/((x - a)(x + a)) dx`

= `1/(2a) int (1/(x - a) - 1/(x + a)) dx`

= `1/(2a) [int 1/(x - a) dx - int 1/(x + a) dx]`

= `1/(2a) [log|x - a| - log|x + a|] + c`

= `1/(2a) log |(x - a)/(x + a)| + c`

∴ `int 1/(x^2 - a^2) dx = 1/(2a) log |(x - a)/(x + a)| + c`

To find: `int 1/(x^2 - 3) dx`

`int 1/(x^2 - 3) dx = int 1/(x^2 - (sqrt3)^2) dx = 1/(2sqrt3) log |(x - sqrt3)/(x + sqrt3)| + c`