Question

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

Answer 1

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

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

= A (2x - 2) +B              ....(i)

Comparing the coefficient of x in (i), we get

1 = 2A

⇒ `A = 1/2`

Comparing the constant terms in (i), we get 

3 = B - 2A

⇒ 3 = B - 1

⇒ B = 4

`I = int (1/2 (2x - 2) + 4)/(x^2 - 2x - 5)`

`1/2 int (2x - 2)/ (x^2 - 2x - 5) dx + 4 int dx /(x^2 - 2x - 5)`

Let `I = 1/2 I_1 + 4I_2`                      ....(ii)

Where `I_1 = int (2x - 2)/ (x^2 - 2x -5)  dx`

Put x² - 2x - 5 = t

⇒ (2x - 2) dx = dt

∴ `I_1 = intdt/t = log |t| = log |x^2 - 2x - 5| + C_1`          ....(iii)

and `I_2 = int dx/ (x^2 - 2x - 5)`

`= dx/ ((x - 1)^2 - 6)`

`= int dx/ ((x - 1)^2 - (sqrt6)^2) = 1/(2sqrt6) log |(x - 1 - sqrt6)/(x - 1 + sqrt6)| + C_2`

Hence from (ii), (iii) and (iv), we get

∴ `I = 1/2 log |(x^2 - 2x - 5)| + 2/ sqrt6 log |(x - 1 - sqrt6)/(x - 1 + sqrt6)| + C`