Key Concepts and Formulas

• Linear Decomposition of Integrands: If the integrand can be expressed as a linear combination of simpler functions, the integral can be split into a sum of integrals.
• Derivative of $\ln(f(x))$: $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$. This is useful when the numerator is the derivative of the denominator (or a constant multiple thereof).
• Indefinite Integral Properties: $\int (af(x) + bg(x)) dx = a\int f(x) dx + b\int g(x) dx$, where $a$ and $b$ are constants.

Step-by-Step Solution

Step 1: Express the integrand as a linear combination.

Our goal is to express the given integrand in a form that allows us to integrate it easily. We want to find constants $A$ and $B$ such that: $2e^x + 3e^{-x} = A(4e^x + 7e^{-x}) + B(4e^x - 7e^{-x})$ This allows us to separate the integral into two parts: one that is a constant and one that is of the form $\frac{f'(x)}{f(x)}$.

Step 2: Solve for A and B.

Expanding the equation from Step 1, we get: $2e^x + 3e^{-x} = (4A + 4B)e^x + (7A - 7B)e^{-x}$ Equating the coefficients of $e^x$ and $e^{-x}$ on both sides, we get the following system of equations: $4A + 4B = 2$ $7A - 7B = 3$ Dividing the first equation by 4 and the second by 7, we have: $A + B = \frac{1}{2}$ $A - B = \frac{3}{7}$ Adding the two equations gives: $2A = \frac{1}{2} + \frac{3}{7} = \frac{7 + 6}{14} = \frac{13}{14}$ Therefore, $A = \frac{13}{28}$. Subtracting the second equation from the first gives: $2B = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14}$ Therefore, $B = \frac{1}{28}$.

Step 3: Rewrite the integral.

Substitute the values of A and B back into the integral: $\int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} dx = \int \frac{\frac{13}{28}(4e^x + 7e^{-x}) + \frac{1}{28}(4e^x - 7e^{-x})}{4e^x + 7e^{-x}} dx$ $= \int \left(\frac{13}{28} + \frac{1}{28}\frac{4e^x - 7e^{-x}}{4e^x + 7e^{-x}}\right) dx$

Step 4: Evaluate the integral.

Now we can split the integral and evaluate each part: $\int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} dx = \frac{13}{28}\int dx + \frac{1}{28}\int \frac{4e^x - 7e^{-x}}{4e^x + 7e^{-x}} dx$ The first integral is simply $\int dx = x$. For the second integral, notice that the numerator is the derivative of the denominator. Let $f(x) = 4e^x + 7e^{-x}$. Then $f'(x) = 4e^x - 7e^{-x}$. Therefore, $\int \frac{4e^x - 7e^{-x}}{4e^x + 7e^{-x}} dx = \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C = \ln|4e^x + 7e^{-x}| + C$ So, the integral becomes: $\int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} dx = \frac{13}{28}x + \frac{1}{28}\ln|4e^x + 7e^{-x}| + C$

Step 5: Compare with the given form and find u and v.

We are given that: $\int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} dx = \frac{1}{14}(ux + v\ln_e(4e^x + 7e^{-x})) + C$ Comparing this with our result, we have: $\frac{13}{28}x + \frac{1}{28}\ln|4e^x + 7e^{-x}| + C = \frac{1}{14}(ux + v\ln_e(4e^x + 7e^{-x})) + C$ Multiplying both sides by 14, we get: $\frac{13}{2}x + \frac{1}{2}\ln|4e^x + 7e^{-x}| + C = ux + v\ln_e(4e^x + 7e^{-x}) + C$ Thus, $u = \frac{13}{2}$ and $v = \frac{1}{2}$.

Step 6: Calculate u + v.

$u + v = \frac{13}{2} + \frac{1}{2} = \frac{14}{2} = 7$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with the signs when equating coefficients and solving for A and B. A small sign error will propagate through the rest of the solution.
• Incorrect Integration: Make sure you correctly identify the form $\frac{f'(x)}{f(x)}$ when integrating. The integral of this form is $\ln|f(x)|$.
• Coefficient Matching: Ensure that you are comparing the coefficients correctly after evaluating the integral and comparing it with the given form.

Summary

We decomposed the integrand into a linear combination of two terms: a constant and a term of the form $\frac{f'(x)}{f(x)}$. We then solved for the constants A and B, integrated each term separately, and compared the result with the given form to find the values of u and v. Finally, we calculated u + v, which equals 7.

Final Answer

The final answer is \boxed{7}.