Question

Find: `int(x+2)/sqrt(9x-x^2)dx`

Answer 1

Let I = `int(x+2)/sqrt(9x-x^2)dx`

I = `int(x+2)/(sqrt(-x^2-9x))dx`

= `int(x+2)/(sqrt(-[x^2-9x+(9/2)^2-(9/2)^2]))dx`

= `int(x+2)/sqrt(-[(x-9/2)^2-81/4])dx`

= `int(x+2)/sqrt(81/4-(x-9/2)^2)dx`

Let t = x – `9/2`

x = t + `9/2`

Add 2 to both sides,

x + 2 = t + `9/2` + 2

⇒ x + 2 = t + `13/2`

Diff. both sides w.r.t ‘x’

⇒ dx = dt

I = `int((t+13/2))/sqrt(81/4-t^2)dt`

= `intt/sqrt(81/4-t^2)dt+13/2intdt/sqrt81/4-t^2`

I₁ = `intt/(sqrt(81/4-t^2))dt`

Let `81/4-t^2 = u`

–2tdt = du

⇒ tdt = `(-du)/2`

= `(-1)/2int(du)/sqrt4`

= `(-1)/2int4^((-1)/2)du`

= `(-1)/2 4^((-1)/2+1)/((-1)/2+1)+C`

= `-sqrt4+C`

= `-sqrt(81/4-t^2)+C`

Substitute t = `x-9/2` 

= `-sqrt(81/4-(x-9/2)^2)+C_1`

= `-sqrt(81/4-x^2-81/4+2*x*9/2)+C_1`

= `-sqrt(-x^2+9x)+C_1`

I₁ = `-sqrt(9x-x^2)+C_1`  ...(1)

I₂ = `13/2intdt/(sqrt(81/4-t^2))`

= `13/2intdt/sqrt((9/2)^2-t^2)`

`intdx/(sqrt(a^2-x^2))=sin^-1  x/a+C`

= `13/2sin^-1  t/(9/2)+C_2`

Substitute, t = `x-9/2`

= `13/2sin^-1  ((x-9/2))/(9/2)+C_2`

`13/2sin^-1  (((2x-9)/2))/(9/2)+C_2`

I₂ = `13/2sin^-1((2x-9)/9)+C_2`   ...(2)

Now by adding (1) and (2)

I = I₁ + I₂

= `-sqrt(9x-x^2)+13/2sin^-1((2x-9)/9)+C`  ...[∵ C = C₁ + C₂]