Question

Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`

Answer 1

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

`5x - 2 = A  d/dx (1 + 2x + 3x^2) + B`

= 5x - 2 = A (6x + 2) + B            ....(i)

Comparing coefficients of x in (i), we get

6A = 5

`therefore A = 5/6`

-2 = 2A + B

`therefore B = -2 - 2A`

`= -2 -2 (5/6)`

`= -2 - 5/3`

`= - 11/3`

`therefore I = int (5/6 (2 + 6x) - 11/3)/(1 + 2x + 3x^2)  dx`

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

`-11/3 int dx/(1 + 2x + 3x^2)  dx`

` Let  I = 5/6 I_1 - 11/3 I_2`                           .... (ii)

`therefore I_1 = 5/6 int (2 + 6x)/(1 + 2x + 3x^2)  dx`

Put 1 + 2x + 3x² = t

(2 + 6x) dx = dt

∴ `I_1 = int dt/t`

`= log |t| + C_1`

`= log |1 + 2x + 3x^2| + C_1`                          .....(iii)

`I_2 = int dx/(1 + 2x + 3x^2)`

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

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

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

`= 1/3 int dx/((x + 1/3)^2 + sqrt2/3)`

`= 1/3 . 1/(sqrt2/3)  tan^-1 ((x + 1/3)/(sqrt2/3))`

`= 1/sqrt2  tan^-1 ((3x + 1)/sqrt2)`              ... (iv)

From (ii), (iii) and (iv), we get

`I = 5/6  log (1 + 2x + 3x^2) - 11/3 xx 1/sqrt2  tan^-1 ((3x + 1)/sqrt2)`

`= 5/6  log (1 + 2x + 3x^2) - 11/(3 sqrt2)  tan^-1 ((3x + 1)/sqrt2) + C`