Answer 1

Consider the integral, `int dx/(x^2 + 4x + 8)`

I = `int 1/(x^2 + 4x + 8)` dx

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

= `int 1/((x + 2)^2 + 2^2) dx`           {Use Intergral Formula : `int 1/(x^2 + a^2) dx = 1/a tan^(-1) (x/a)}`

`= 1/2 tan^(-1)  (x + 2)/2 + C`