Question

Find: 

`int (3x+5)/sqrt (x^2 + 2x +4)  dx`

Answer 1

`int (3x+5)/sqrt (x^2 + 2x +4)  dx`

Let `I int (3x+5)/sqrt (x^2 + 2x +4)`

`= int (3(x+1))/(x^2+2x+4) + 2/sqrt ((x+1)^2+(sqrt 3)^2)`

`I_1 = int (3x+5)/sqrt ((x+1)^2+(sqrt 3)^2)`

I = I₁ + I₂

`I = 3int (x+1)/sqrt (x^2+2x+4)`

`I_1 = 3/2intdt/sqrtt`

Let x² + 2x + 4 = t

(2x + 2) dx = dt

(x + 1) dx = `dt/2`

`I_1 = 3/2intt^(-1/2) dt`

`I_1 = 3/2 t^(1/2)/(1/2)`

`I_1 = 3 sqrt (x^2 +2x+4)`

`I_2 = 2int 1/(sqrt((x+1)^2+(sqrt3)^2))`

`I_2 = 2log |x+1+sqrt(x^2+2x+4)|`

`I = 3sqrt(x^2+2x+4) + 2log |x+1+sqrt(x^2+2x+4)| + C`