If `int(3e^(x)+4e^(-x))/(5e^(x)+6e^(-x))=ax+b log(5e^(2x)+6)+c,` then

Options

A. `a=(2)/(3), b=-30`
B. `a=(3)/(2), b=(-1)/(30)`
C. `a=(2)/(3), b=(-1)/(30)`
D. `a=-(3)/(2), b=-30`

Answer

To solve the integral

\[ \int \frac{3e^{x} + 4e^{-x}}{5e^{x} + 6e^{-x}} \, dx = ax + b \log(5e^{2x} + 6) + c, \]

we will differentiate both sides and compare coefficients to find the values of \(a\) and \(b\).

### Step 1: Differentiate both sides

Differentiating the left-hand side:

\[ \frac{d}{dx} \left( \int \frac{3e^{x} + 4e^{-x}}{5e^{x} + 6e^{-x}} \, dx \right) = \frac{3e^{x} + 4e^{-x}}{5e^{x} + 6e^{-x}}. \]

Now differentiate the right-hand side:

\[ \frac{d}{dx} \left( ax + b \log(5e^{2x} + 6) + c \right) = a + b \cdot \frac{1}{5e^{2x} + 6} \cdot \frac{d}{dx}(5e^{2x} + 6). \]

### Step 2: Differentiate \(5e^{2x} + 6\)

Using the chain rule:

\[ \frac{d}{dx}(5e^{2x} + 6) = 5 \cdot 2e^{2x} = 10e^{2x}. \]

So, we have:

\[ \frac{d}{dx} \left( ax + b \log(5e^{2x} + 6) + c \right) = a + b \cdot \frac{10e^{2x}}{5e^{2x} + 6}. \]

### Step 3: Set the derivatives equal

Equating the derivatives from both sides:

\[ \frac{3e^{x} + 4e^{-x}}{5e^{x} + 6e^{-x}} = a + \frac{10be^{2x}}{5e^{2x} + 6}. \]

### Step 4: Find a common denominator

To compare coefficients, we can rewrite the left-hand side:

Let’s multiply the numerator and denominator of the left side by \(e^{x}\):

\[ \frac{3e^{2x} + 4}{5e^{2x} + 6}. \]

### Step 5: Compare coefficients

Now we can compare coefficients of \(e^{2x}\) and the constant terms:

1. Coefficient of \(e^{2x}\):

From the left side: \(3\)

From the right side: \(10b\)

Thus, we have:

\[ 10b = 3 \implies b = \frac{3}{10}. \]

2. Constant terms:

From the left side: \(4\)

From the right side: \(a \cdot (5e^{2x} + 6)\) evaluated at \(e^{2x} = 0\) gives \(6a\).

Thus, we have:

\[ 6a = 4 \implies a = \frac{4}{6} = \frac{2}{3}. \]

### Final Values

Thus, we find:

\[ a = \frac{2}{3}, \quad b = \frac{3}{10}. \]