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Step-by-Step Solution
Step 1: Identify the integrand
We need to evaluate
\displaystyle \int \frac{2 e^x + 3 e^{-x}}{4 e^x + 7 e^{-x}} \, dx.
The problem states that this integral can be written in the form
\displaystyle \frac{1}{14}\bigl(u x + v \ln_e(4 e^x + 7 e^{-x})\bigr) + C,
and we want to find the value of u + v.
Step 2: Rewrite the numerator
To integrate, we try to express
\displaystyle 2 e^x + 3 e^{-x}
as a linear combination of
\displaystyle (4 e^x + 7 e^{-x})
and its derivative. Notice that the derivative of the denominator
\displaystyle (4 e^x + 7 e^{-x})
is
\displaystyle 4 e^x - 7 e^{-x}.
Thus, let us set up the following decomposition:
\displaystyle
2 e^x + 3 e^{-x} \;=\; A \bigl(4 e^x + 7 e^{-x}\bigr) \;+\; B \bigl(4 e^x - 7 e^{-x}\bigr).
Step 3: Solve for the constants A and B
Collect like terms in powers of e^x and e^{-x} .
1) Coefficient of e^x :
\displaystyle 2 = 4A + 4B.
2) Coefficient of e^{-x} :
\displaystyle 3 = 7A - 7B.
Adding and subtracting these equations helps us find A and B .
From
\displaystyle 2 = 4(A + B) \implies A + B = \frac{1}{2}.
From
\displaystyle 3 = 7(A - B) \implies A - B = \frac{3}{7}.
Solving, we get:
\displaystyle
A = \frac{1}{2}\left(\frac{1}{2} + \frac{3}{7}\right)
= \frac{1}{2}\left(\frac{7 + 6}{14}\right)
= \frac{13}{28},
\quad
B = A - \frac{3}{7}
= \frac{13}{28} - \frac{12}{28}
= \frac{1}{28}.
Step 4: Rewrite the integral using A and B
Hence,
\displaystyle
2 e^x + 3 e^{-x}
= \frac{13}{28}\bigl(4 e^x + 7 e^{-x}\bigr) + \frac{1}{28}\bigl(4 e^x - 7 e^{-x}\bigr).
Thus,
\displaystyle
\frac{2 e^x + 3 e^{-x}}{4 e^x + 7 e^{-x}}
= \frac{13}{28} + \frac{1}{28}\,\frac{4 e^x - 7 e^{-x}}{4 e^x + 7 e^{-x}}.
Step 5: Integrate term by term
Now the integral becomes:
\displaystyle
\int \frac{2 e^x + 3 e^{-x}}{4 e^x + 7 e^{-x}} \, dx
= \int \left(\frac{13}{28} + \frac{1}{28} \cdot \frac{4 e^x - 7 e^{-x}}{4 e^x + 7 e^{-x}}\right)\, dx.
Split this into two integrals:
\displaystyle
\int \frac{13}{28} \, dx
= \frac{13}{28}\, x.
The second part:
\displaystyle
\frac{1}{28} \int \frac{4 e^x - 7 e^{-x}}{4 e^x + 7 e^{-x}} \, dx.
We observe that 4 e^x - 7 e^{-x} is exactly the derivative of 4 e^x + 7 e^{-x} . Hence:
\displaystyle
\frac{1}{28} \int \frac{4 e^x - 7 e^{-x}}{4 e^x + 7 e^{-x}} \, dx
= \frac{1}{28} \ln\bigl|4 e^x + 7 e^{-x}\bigr| + C.
Therefore, the integral evaluates to:
\displaystyle
\int \frac{2 e^x + 3 e^{-x}}{4 e^x + 7 e^{-x}} \, dx
= \frac{13}{28} x + \frac{1}{28} \ln\bigl|4 e^x + 7 e^{-x}\bigr| + C.
Step 6: Compare with the given form and find u + v
The expression is given as
\displaystyle \frac{1}{14} \bigl(u x + v \ln(4 e^x + 7 e^{-x})\bigr) + C.
We match coefficients as follows:
\displaystyle
\frac{13}{28} x
= \frac{1}{14} \cdot u x
\quad \Longrightarrow \quad
u = \frac{13}{2}.
\displaystyle
\frac{1}{28} \ln\bigl|4 e^x + 7 e^{-x}\bigr|
= \frac{1}{14}\, v \ln\bigl|4 e^x + 7 e^{-x}\bigr|
\quad \Longrightarrow \quad
v = \frac{1}{2}.
Hence:
\displaystyle
u + v
= \frac{13}{2} + \frac{1}{2}
= 7.
Final Answer
The required sum \displaystyle u + v is \boxed{7}.
