Question

Integrate the following with respect to x:

`(2x + 1)/sqrt(9 + 4x - x^2)`

Answer 1

Let 2x + 1 = `"A" "d"/("d"x) (9 + 4x - x^2) + "B"`

2x + 1 = A(4 – 2x) + B

2x + 1 = 4A – 2Ax + B

– 2A = 2

⇒ A = – 1

4A + B = 1

⇒ 4(– 1) + B = 1

B = 1 + 4 = 5

B = 5

2x + 1 = – 1(4 – 2x) + 5

`int (2x + 1)/sqrt(9 + 4x - x^2)  "d"x = int (-(4 - 2x) + 5)/sqrt(9 + 4x - x^2)`

= `int (- (4 - 2x))/sqrt(9 + 4x - x^2)  "d"x + 5 int ("d"x)/sqrt(9 + 4x - x^2)`

Put 9 + 4x – x² = t

(4x – 2) dx = dt

= `int (- "dt")/sqrt("t") + 5 int ("d"x)/sqrt(9 - (x^2 - 4x))`

= `int - "t"^(- 1/2)  "dt" + 5 int ("d"x)/sqrt(9 - [(x - 2)^2 - 2^2]`

= `- ("t"^(- 1/2 + 1))/(- 1/2 + 1) + 5 int ("d"x)/sqrt(9 - [(x - 2)^2 - 2^2]`

= `- ("t"^(1/2))/(1/2) + 5 int ("d"x)/sqrt(9 - [(x - 2)^2 - 4]`

= ` - 2sqrt("t") + 5 int ("d"x)/sqrt(13 - (x - 2)^2`

Put x – 2 = u

dx = du

= `- 2 sqrt(9 + 4x - x^2) + 5 int ("d"x)/sqrt((sqrt(13))^2 - "u"^2)`

= `- 2 sqrt(9 + 4x - x^2) + 5 sin^-1 ("u"/sqrt(13)) + "c"`

`int (2x + 1)/sqrt(9 + 4x - x^2)  "d"x = - 2sqrt(9 + 4x - x^2) + 5 sin^-1 ((x - 2)/sqrt(13)) + "c"`