Question

Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`

Answer 1

Let, I `= int (6x + 7)/sqrt((x - 5)(x - 4))  dx`

`= int (6x + 7)/sqrt(x^2 - 9x + 20)  dx`

`6x + 7 = A  d/dx (x^2 - 9x + 20) + B`

= A (2x - 9) + B                       ....(i)

Comparing coefficient of x in (i), we get

6 = 2A

`therefore A = 3`

7 = - 9A + B

= - 27 + B

`therefore B = 34`

`therefore I = int (3 (2x - 9) + 34)/sqrt(x^2 - 9x + 20) dx`

`I = 3= int ((2x - 9))/(x^2 - 9x + 20)  dx + 34  int dx/sqrt(x^2 - 9x + 20)`

`= 3I_1 - 34  I_2`                                ....(ii)

`therefore I_1 = int ((2x - 9))/sqrt(x^2 - 9x + 20)  dx`

Put x² - 9x + 20 = t

(2x - 9) dx = dt

∴ `I_1 = int dt/sqrtt`

`therefore dt/sqrtt int t^(1/2)  dt = (t^(1/2)/(1/2)) + C_1`

`= 2 sqrtt + C_1` 

`2 = sqrt(x^2 - 9x + 20) + C_1`                    .....(iii)

`I_2 = int dx/sqrt(x^2 - 9x + 20)`

`= int dx/sqrt (x^2 - 9x + (9/2)^2 - (9/2)^2 + 20)`

`= dx/ sqrt ((x = 9/2)^2 - 81/4 + 20)`

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

`= log |(x - 9/2) + sqrt ((x - 9/2)^2 - (1/2)^2)| + C_2`

`= log |(x - 9/2) + sqrt (x^2 - 9x + 20)| + C_2`              ....(iv)

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

`I = 3 xx2 sqrt (x^2 - 9x + 20) + 34  log |(x - 9/2) + sqrt (x^2 - 9x + 20)| + C`

or `I = 6 sqrt (x^2 - 9x + 20) + 34 log |(x - 9/2) + sqrt (x^2 - 9x + 20)| + C`