Key Concepts and Formulas

• Linear Decomposition of Integrands: Expressing a complex rational function as a sum of simpler functions to facilitate integration.
• Standard Integrals:
  • $\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C$
  • $\int \sqrt{x^2 + a^2} dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln|x + \sqrt{x^2 + a^2}| + C$
• Completing the Square: Transforming a quadratic expression into a perfect square plus a constant.

Step-by-Step Solution

Step 1: Decompose the integrand.

We aim to express the numerator in terms of the derivative and the original quadratic inside the square root. Let $2x^2 + 5x + 9 = A(x^2 + x + 1) + B(2x + 1) + C$ Expanding the right side, we have $2x^2 + 5x + 9 = Ax^2 + Ax + A + 2Bx + B + C$ $2x^2 + 5x + 9 = Ax^2 + (A + 2B)x + (A + B + C)$ Comparing coefficients, we get the following system of equations: \begin{align*} A &= 2 \ A + 2B &= 5 \ A + B + C &= 9 \end{align*} From the first equation, $A = 2$. Substituting into the second equation, $2 + 2B = 5$, so $2B = 3$, and $B = \frac{3}{2}$. Substituting into the third equation, $2 + \frac{3}{2} + C = 9$, so $C = 9 - 2 - \frac{3}{2} = 7 - \frac{3}{2} = \frac{11}{2}$. Therefore, $2x^2 + 5x + 9 = 2(x^2 + x + 1) + \frac{3}{2}(2x + 1) + \frac{11}{2}$

Step 2: Rewrite the integral.

Using the decomposition from Step 1, we can rewrite the integral as: $\int \frac{2x^2 + 5x + 9}{\sqrt{x^2 + x + 1}} dx = \int \frac{2(x^2 + x + 1) + \frac{3}{2}(2x + 1) + \frac{11}{2}}{\sqrt{x^2 + x + 1}} dx$ $= 2 \int \frac{x^2 + x + 1}{\sqrt{x^2 + x + 1}} dx + \frac{3}{2} \int \frac{2x + 1}{\sqrt{x^2 + x + 1}} dx + \frac{11}{2} \int \frac{1}{\sqrt{x^2 + x + 1}} dx$ $= 2 \int \sqrt{x^2 + x + 1} dx + \frac{3}{2} \int \frac{2x + 1}{\sqrt{x^2 + x + 1}} dx + \frac{11}{2} \int \frac{1}{\sqrt{x^2 + x + 1}} dx$

Step 3: Evaluate the second integral.

Let $u = x^2 + x + 1$. Then $du = (2x + 1) dx$. So, $\frac{3}{2} \int \frac{2x + 1}{\sqrt{x^2 + x + 1}} dx = \frac{3}{2} \int \frac{du}{\sqrt{u}} = \frac{3}{2} \int u^{-1/2} du = \frac{3}{2} \cdot 2 u^{1/2} + C_1 = 3 \sqrt{x^2 + x + 1} + C_1$

Step 4: Evaluate the third integral.

Complete the square in the denominator: $x^2 + x + 1 = x^2 + x + \frac{1}{4} + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$ So, $\frac{11}{2} \int \frac{1}{\sqrt{x^2 + x + 1}} dx = \frac{11}{2} \int \frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}} dx$ Using the standard integral $\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C$, we have $\frac{11}{2} \int \frac{dx}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}} = \frac{11}{2} \ln\left|x + \frac{1}{2} + \sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\right| + C_2$ $= \frac{11}{2} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C_2$

Step 5: Evaluate the first integral.

$2 \int \sqrt{x^2 + x + 1} dx = 2 \int \sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dx$ Using the standard integral $\int \sqrt{x^2 + a^2} dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln|x + \sqrt{x^2 + a^2}| + C$, we have $2 \left[ \frac{x + \frac{1}{2}}{2} \sqrt{x^2 + x + 1} + \frac{(\sqrt{3}/2)^2}{2} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| \right] + C_3$ $= 2 \left[ \frac{2x + 1}{4} \sqrt{x^2 + x + 1} + \frac{3/4}{2} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| \right] + C_3$ $= \frac{2x + 1}{2} \sqrt{x^2 + x + 1} + \frac{3}{4} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C_3$ $= x \sqrt{x^2 + x + 1} + \frac{1}{2} \sqrt{x^2 + x + 1} + \frac{3}{4} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C_3$

Step 6: Combine the results.

$\int \frac{2x^2 + 5x + 9}{\sqrt{x^2 + x + 1}} dx = x \sqrt{x^2 + x + 1} + \frac{1}{2} \sqrt{x^2 + x + 1} + \frac{3}{4} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + 3 \sqrt{x^2 + x + 1} + \frac{11}{2} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C$ $= x \sqrt{x^2 + x + 1} + \left(\frac{1}{2} + 3\right) \sqrt{x^2 + x + 1} + \left(\frac{3}{4} + \frac{22}{4}\right) \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C$ $= x \sqrt{x^2 + x + 1} + \frac{7}{2} \sqrt{x^2 + x + 1} + \frac{25}{4} \ln\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C$

Step 7: Compare with the given form and find $\alpha$ and $\beta$.

Comparing this with the given form $x \sqrt{x^2 + x + 1} + \alpha \sqrt{x^2 + x + 1} + \beta \log_e\left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C$ we have $\alpha = \frac{7}{2}$ and $\beta = \frac{25}{4}$.

Step 8: Calculate $\alpha + 2\beta$.

$\alpha + 2\beta = \frac{7}{2} + 2\left(\frac{25}{4}\right) = \frac{7}{2} + \frac{25}{2} = \frac{32}{2} = 16$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when completing the square and substituting back.
• Incorrect Standard Integrals: Make sure you correctly remember the standard integral formulas.
• Simplification: Always simplify your expression as much as possible after each integration step.

Summary

We decomposed the integrand into simpler terms using linear combinations of the quadratic expression inside the square root and its derivative. We then evaluated each integral separately using standard integral formulas and completing the square where necessary. Finally, we combined the results and compared them with the given form to find the values of $\alpha$ and $\beta$, and calculated $\alpha + 2\beta$.

The final answer is \boxed{16}.