Question

Evaluate :

`∫(x+2)/sqrt(x^2+5x+6)dx`

Answer 1

`I=∫(x+2)/sqrt(x^2+5x+6)dx `

Multiplying and dividing by 2, we get

`I=1/2∫(2x+4)/sqrt(x^2+5x+6)dx `

Adding and subtracting 1 to the numerator, we get:

`I=1/2∫(2x+4+1-1)/sqrt(x^2+5x+6)dx`

` I=1/2∫(2x+5)/sqrt(x2+5x+6)dx -1/2∫1/sqrt(x^2+5x+6)dx`

`"Let" I_1=1/2∫(2x+5)/sqrt(x^2+5x+6)dx `

Put x²+5x+6=t

Differentiating with respect to x, we get:

(2x+5)dx=dt

`I_1=intdt/sqrtt`

`I_1=2sqrtt+c`

`I_1=2sqrt(x^2+5x+6)+c`

`1/2 int "dt"/sqrt t =∫1/sqrt(x^2+5x+(5/2)^2-(5/2)^2+6)dx`

`1/2 int "dt"/sqrt t - 1/2 int "dx"/sqrt(x^2+5x+6 + (5/2)^2 - 25/4)dx`

`1/2 "t"^(1/2)/(1/2) int 1/sqrt((x+5/2)^2-(1/2)^2)dx`

`= 1/2 xx 2 xx "t"^(1/2) - 1/2 |"log" x + 5/2 + sqrt (x^2 + 5x + 6)| + "C"`

`= sqrt (x^2 + 5x + 6) - 1/2 "log" |sqrt (x^2 + 5x + 6)| + "C"`