Question

Integrate the following with respect to x:

`(2x - 3)/(x^2 + 4x - 12)`

Answer 1

Let 2x – 3 = `"A" "d"/("d"x) (x^2 + 4x - 12) + "B"`

2x – 3 = A(2x + 4) + B

2x – 3 = 2Ax + 4A + B

2A = 2

⇒ A = 1

4A + B = – 3

4 × 1 + B – 3

⇒ B = – 3 – 4 = – 7

2x – 3 = 1(2x + 4) – 7

`int (2x - 3)/(x^2 + 4x - 12)  "d"x = int (2x + 4 - 7)/(x^2 + 4x - 12)  "d"x`

= `int (2x + 4)/(x^2 + 4x - 12)  "d"x - 7 int ("d"x)/(x^2 + 4x - 12)`

Put x² + 4x – 12 = t

(2x + 4)dx = dt

= `int "dt"/"t" - 7 int ("d"x)/((x + 2)^2 - 2^2 - 12)`

= `log |"t"| - 7 int ("d"x)/((x + 2)^2 - 4 - 12)`

= `log |x^2 + 4x - 12| - 7 int ("d"x)/((x + 2)^2 - 16)`

= `log |x^2 + 4x - 12| - 7 int ("d"x)/((x + 2)^2 - 4^2)`

Put x + 2 = u

dx = du

= `log |x^2 + 4x - 12| - 7 int "du"/("u"^2 - 4^2)`

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

= `log |x^2 + 4x - 12| - 7 xx 1/(2 xx 4) log |("u" - 4)/("u" + 4)| + "c"`

= `log |x^2 + 4x - 12| - 7/8 xx log |(x + 2 - 4)/(x + 2 - 4)| + "c"`

`int (2x - 3)/(x^2 + 4x - 12)  "d"x = log |x^2 + 4x - 12| - 7/8 xx log |(x - 2)/(x + 6)| + "c"`