1. **State the problem:** We need to evaluate the integral $$I = \int \frac{dx}{\sqrt{2x^2 + 3x + 4}}.$$\n\n2. **Identify the integral type:** This is an integral of the form $$\int \frac{dx}{\sqrt{ax^2 + bx + c}}$$ which often requires completing the square in the quadratic expression under the square root.\n\n3. **Complete the square:** For the quadratic $$2x^2 + 3x + 4,$$ factor out the coefficient of $$x^2$$ from the first two terms:\n$$2x^2 + 3x + 4 = 2\left(x^2 + \frac{3}{2}x\right) + 4.$$\nComplete the square inside the parentheses:\n$$x^2 + \frac{3}{2}x = \left(x + \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 = \left(x + \frac{3}{4}\right)^2 - \frac{9}{16}.$$\nSubstitute back:\n$$2\left(x + \frac{3}{4}\right)^2 - 2 \cdot \frac{9}{16} + 4 = 2\left(x + \frac{3}{4}\right)^2 - \frac{9}{8} + 4 = 2\left(x + \frac{3}{4}\right)^2 + \frac{23}{8}.$$\n\n4. **Rewrite the integral:**\n$$I = \int \frac{dx}{\sqrt{2\left(x + \frac{3}{4}\right)^2 + \frac{23}{8}}}.$$\n\n5. **Substitute:** Let $$t = x + \frac{3}{4},$$ so $$dx = dt.$$\nThe integral becomes:\n$$I = \int \frac{dt}{\sqrt{2t^2 + \frac{23}{8}}} = \int \frac{dt}{\sqrt{2t^2 + \frac{23}{8}}}.$$\n\n6. **Factor out 2 inside the square root:**\n$$\sqrt{2t^2 + \frac{23}{8}} = \sqrt{2\left(t^2 + \frac{23}{16}\right)} = \sqrt{2} \sqrt{t^2 + \frac{23}{16}}.$$\nSo the integral is:\n$$I = \int \frac{dt}{\sqrt{2} \sqrt{t^2 + \frac{23}{16}}} = \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{t^2 + \left(\frac{\sqrt{23}}{4}\right)^2}}.$$\n\n7. **Use the standard integral formula:**\n$$\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C.$$\nHere, $$a = \frac{\sqrt{23}}{4}.$$\n\n8. **Apply the formula:**\n$$I = \frac{1}{\sqrt{2}} \ln\left|t + \sqrt{t^2 + \left(\frac{\sqrt{23}}{4}\right)^2}\right| + C = \frac{1}{\sqrt{2}} \ln\left|x + \frac{3}{4} + \sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{23}{16}}\right| + C.$$\n\n**Final answer:**\n$$I = \frac{1}{\sqrt{2}} \ln\left|x + \frac{3}{4} + \sqrt{2x^2 + 3x + 4 \over 2}\right| + C.$$