$$\int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}} dx = \frac{1}{14}(ux + v\log_e(4e^x + 7e^{-x})) + C$$, where $$C$$ is a constant of integration, then $$u + v$$ is equal to _________.

1. Expressing Numerator in Terms of Denominator and its Derivative

For integrals of the form $$\int \frac{p e^x + q e^{-x}}{r e^x + s e^{-x}} dx$$, we express the numerator as:

$$\text{Numerator} = A(\text{Denominator}) + B\left(\frac{d}{dx}(\text{Denominator})\right)$$

The given integral is:

$$\int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}} dx$$

Here, the denominator is $$4e^x + 7e^{-x}$$, and its derivative is $$4e^x - 7e^{-x}$$.

$$2e^x + 3e^{-x} = A(4e^x + 7e^{-x}) + B(4e^x - 7e^{-x})$$

2. Solving for Constants A and B

Equating the coefficients of $$e^x$$ and $$e^{-x}$$ on both sides:

$$4A + 4B = 2 \implies A + B = \frac{1}{2}$$

$$7A - 7B = 3 \implies A - B = \frac{3}{7}$$

Adding the two equations:

$$2A = \frac{1}{2} + \frac{3}{7} = \frac{13}{14} \implies A = \frac{13}{28}$$

Subtracting the second equation from the first:

$$2B = \frac{1}{2} - \frac{3}{7} = \frac{1}{14} \implies B = \frac{1}{28}$$

3. Evaluating the Integral

Substitute the expression back into the integral:

$$\int \frac{\frac{13}{28}(4e^x + 7e^{-x}) + \frac{1}{28}(4e^x - 7e^{-x})}{4e^x + 7e^{-x}} dx$$

$$\int \frac{13}{28} dx + \int \frac{1}{28} \frac{4e^x - 7e^{-x}}{4e^x + 7e^{-x}} dx$$

$$\frac{13}{28}x + \frac{1}{28}\log_e(4e^x + 7e^{-x}) + C$$

4. Finding u, v, and u + v

Taking $$\frac{1}{14}$$ common out of the result to match the given form:

$$\frac{1}{14}\left(\frac{13}{2}x + \frac{1}{2}\log_e(4e^x + 7e^{-x})\right) + C$$

Comparing this with $$\frac{1}{14}(ux + v\log_e(4e^x + 7e^{-x})) + C$$:

$$u = \frac{13}{2}$$

$$v = \frac{1}{2}$$

Calculating $$u + v$$:

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

Answer:

The value of $$u + v$$ is $$7$$.