Find: ∫ (3x+5)/(√(x^2+2x+4)) dx.

The correct answer is Option (1) → $3\sqrt{x^2+2x+4} + 2\ln|x+1 + \sqrt{x^2+2x+4}| + C$

Let $I = \int \frac{3x+5}{\sqrt{x^2+2x+4}} dx$

$= \int \left[ \frac{3(x+1)}{\sqrt{x^2+2x+4}} + \frac{2}{\sqrt{(x+1)^2+(\sqrt{3})^2}} \right] dx$

$I = I_1 + I_2$

where, $I_1 = 3 \int \frac{x+1}{\sqrt{x^2+2x+4}} dx$

Let $x^2+2x+4 = t$

$(2x+2) dx = dt$

$(x+1) dx = \frac{dt}{2}$

$I_1 = \frac{3}{2} \int t^{-1/2} dt$

$I_1= \frac{3}{2} \frac{t^{1/2}}{1/2} + C_1$

$∴I_1= 3\sqrt{x^2+2x+4} + C_1$

and $I_2 = 2 \int \frac{1}{\sqrt{(x+1)^2+(\sqrt{3})^2}} dx$

$I_2 = 2 \log |x+1 + \sqrt{x^2+2x+4}| + C_2$

Thus, $I = 3\sqrt{x^2+2x+4} + 2\log |x+1 + \sqrt{x^2+2x+4}| + C$ where $C = C_1 + C_2$