Question

Integrate the function `1/(9x^2 + 6x + 5)`

Answer 1

`1/(9x^2 + 6x + 5)`

I = `int1/(9x^2 + 6x + 5)dx = int 1/((3x + 1)^2 + 2^2)`

let (3x + 1) = t ⇒ 3dx = dt

I = `int 1/((3x + 1)^2 + 2^2)dx`

= `1/3 int 1/(t^2 + 2^2)dt`

= `1/3[1/2tan^-1 (t/2)] + C`

= `1/6 tan^-1 (t/2) + C`

Substituting the value of t

= `1/6 tan^-1((3x + 1)/2) + C`

Answer 2

Let `I = int dx/ (9x^2 + 6x + 5)`

`= 1/9 int dx/ (x^2 + 2/3x + 5/9)`

`= 1/9 int dx/ ((x^2 + 2/3x + 1/9) + (5/9 - 1/9))`

`= 1/9 int dx/ ((x + 1/3)^2 + (2/3)^2)`

`= 1/9 xx 1/ (2/3) tan^-1 ((x + 1/3)/ (2/3)) + C`             `....[∵ int dx/ (x^2 + a^2) = 1/a tan^-1  x/a + C]`

`= 1/6 tan^-1 ((3x + 1)/2) + C`