Let `f :R to [(3)/(4), oo)` be a surjective quadratic function with line of symmetry `2x -1=0 and f (1) =1`
`int (e ^(x))/(f (e ^(x)))dx`

Options

A. `cot ^(-1) ((2x ^(2) -1)/(sqrt3))+C`
B. `(2)/(sqrt3)cot ^(-1) ((2x ^(2) -1)/(sqrt3))+C`
C. `tan ^(-1) ((2x ^(2) -1)/(sqrt3))+C`
D. `(2)/(sqrt3)tan ^(-1) ((2e^(x) -1)/(sqrt3))+C`

Answer

To solve the given problem step by step, we will follow the procedure outlined in the video transcript. The problem involves finding the integral of the function \( \frac{e^x}{f(e^x)} \), where \( f \) is a surjective quadratic function.

### Step 1: Determine the quadratic function \( f(x) \)

Given the line of symmetry \( 2x - 1 = 0 \), we can find the vertex of the quadratic function. The line of symmetry is given by \( x = \frac{1}{2} \).

Since \( f \) is a surjective quadratic function that takes values from \( \frac{3}{4} \) to \( \infty \), we can express \( f(x) \) in vertex form:

\[ f(x) = a\left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \]

### Step 2: Use the condition \( f(1) = 1 \)

We know that \( f(1) = 1 \). Substituting \( x = 1 \) into the function gives:

\[ f(1) = a\left(1 - \frac{1}{2}\right)^2 + \frac{3}{4} = a\left(\frac{1}{2}\right)^2 + \frac{3}{4} = \frac{a}{4} + \frac{3}{4} \]

Setting this equal to 1:

\[ \frac{a}{4} + \frac{3}{4} = 1 \]

Subtracting \( \frac{3}{4} \) from both sides:

\[ \frac{a}{4} = 1 - \frac{3}{4} = \frac{1}{4} \]

Multiplying both sides by 4:

\[ a = 1 \]

### Step 3: Write the function \( f(x) \)

Substituting \( a = 1 \) back into the function:

\[ f(x) = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \]

Expanding this:

\[ f(x) = \left(x^2 - x + \frac{1}{4}\right) + \frac{3}{4} = x^2 - x + 1 \]

### Step 4: Find \( f(e^x) \)

Now, we need to find \( f(e^x) \):

\[ f(e^x) = (e^x)^2 - e^x + 1 = e^{2x} - e^x + 1 \]

### Step 5: Set up the integral

We need to evaluate the integral:

\[ \int \frac{e^x}{f(e^x)} \, dx = \int \frac{e^x}{e^{2x} - e^x + 1} \, dx \]

### Step 6: Substitute \( t = e^x \)

Let \( t = e^x \), then \( dx = \frac{dt}{t} \). The integral becomes:

\[ \int \frac{t}{t^2 - t + 1} \cdot \frac{dt}{t} = \int \frac{1}{t^2 - t + 1} \, dt \]

### Step 7: Complete the square in the denominator

To integrate \( \frac{1}{t^2 - t + 1} \), we complete the square:

\[ t^2 - t + 1 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4} \]

### Step 8: Perform the integral

Now, we can use the formula for the integral of the form \( \int \frac{1}{x^2 + a^2} \):

\[ \int \frac{1}{\left(t - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dt = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2(t - \frac{1}{2})}{\sqrt{3}}\right) + C \]

### Step 9: Substitute back \( t = e^x \)

Substituting back \( t = e^x \):

\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2(e^x - \frac{1}{2})}{\sqrt{3}}\right) + C \]

### Final Answer

Thus, the final result of the integral is:

\[ \int \frac{e^x}{f(e^x)} \, dx = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2e^x - 1}{\sqrt{3}}\right) + C \]