integration of x dx/(x^2 +4x+7)^1/2?

Question

no trigonometric form

Answer 1

x^2+4x+7 = (x+2)^2 + 3

let x+2 = y

dx = dy
we have to integrate (y-2)dy /(y^2+3)^(1/2)

= ydy/(y^2+3)^(1/20 - 2/(y^2+3)^(1/2)

for the first expression ley y^2+ 3 = t so ydy = dt/2

integral is dt/2(t^(1/2)) = t^^(1/2) = (y^2+3)^(1/2)

for solving the -2dy/(y^2+3)^(1/2) unfortunately one cannot integrate without using trigonometry

let y = sqrt(3) tan t
y^2 + 3 = 3 sec^2 t or sqrt(y^2+3) = sqrt(3) sec t

dy/dt = sqrt(3) sec^2 t
so dy/(y^2+3)^(1/2) = sec t dt

integrate to get ln | sec t + tan t|

now you can put the value and get the result