The answer is not -1.55.

$\begin{aligned} I & = \int_0^2 \frac {\sqrt x+2}{\sqrt x+4} dx \\ & = \int_0^2 \left(1 - \frac 2{\sqrt x+4}\right) dx & \small \color{#3D99F6} \text{Let }u^2 = x \implies 2u \ du = dx \\ & = \int_0^2 dx - \int_0^{\sqrt 2} \frac {4u}{u+4} du \\ & = x \ \bigg|_0^2 - \int_0^{\sqrt 2} \left(4 - \frac {16}{u+4} \right) du \\ & = 2 - \int_0^{\sqrt 2} 4 \ du + \int_0^{\sqrt 2} \frac {16}{u+4} du \\ & = 2 - 4u \ \bigg|_0^{\sqrt 2} + 16\ln(u+4) \ \bigg|_0^{\sqrt 2} \\ & = 2 - 4 \sqrt 2 + 16 (\ln(\sqrt 2+4) - \ln (4)) \\ & \approx \boxed{1.19} \end{aligned}$

Let y = x^(1/2) + 4. Then x^(1/2) + 2 = y - 2, and dx = 2(y - 4)dy. When x = 0, y =4; when x = 2, y = x^(1/2) + 4 The integral becomes 2 [integral (y - 2)(y - 4)/y]dy evaluated from y = 4 to y = 4 + 2^(1/2),, or: [(y^2 - 12 y + 16*ln(y)]. Substituting the upper and lower values, we get 1.186878168. Ed Gray