1. **State the problem:** Evaluate the integral $$\int \frac{dx}{4x^2 + 20x + 26}$$. 2. **Complete the square:** The quadratic in the denominator is $$4x^2 + 20x + 26$$. Factor out 4 from the first two terms: $$4x^2 + 20x + 26 = 4(x^2 + 5x) + 26$$. Complete the square inside the parentheses: $$x^2 + 5x = x^2 + 5x + \left(\frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 = \left(x + \frac{5}{2}\right)^2 - \frac{25}{4}$$. Substitute back: $$4\left(\left(x + \frac{5}{2}\right)^2 - \frac{25}{4}\right) + 26 = 4\left(x + \frac{5}{2}\right)^2 - 25 + 26 = 4\left(x + \frac{5}{2}\right)^2 + 1$$. 3. **Rewrite the integral:** $$\int \frac{dx}{4\left(x + \frac{5}{2}\right)^2 + 1}$$. 4. **Make substitution:** Let $$u = x + \frac{5}{2}$$, so $$du = dx$$. The integral becomes: $$\int \frac{du}{4u^2 + 1}$$. 5. **Factor denominator:** $$4u^2 + 1 = (2u)^2 + 1^2$$. 6. **Use standard integral formula:** $$\int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$$. Here, $$a = \frac{1}{2}$$ because $$4u^2 + 1 = (2u)^2 + 1^2$$, so rewrite as: $$\int \frac{du}{(\frac{1}{2})^2 + u^2}$$. 7. **Apply formula:** $$\int \frac{du}{(\frac{1}{2})^2 + u^2} = \frac{1}{\frac{1}{2}} \arctan\left(\frac{u}{\frac{1}{2}}\right) + C = 2 \arctan(2u) + C$$. 8. **Substitute back:** $$2 \arctan\left(2\left(x + \frac{5}{2}\right)\right) + C = 2 \arctan(2x + 5) + C$$. **Final answer:** $$\int \frac{dx}{4x^2 + 20x + 26} = 2 \arctan(2x + 5) + C$$.