If `f(x)=x^(5)+5x-1` then `int_(5)^(41)(dx)/((f^(-1)(x))^(5)+5f^(-1)(x))` equals

Options

A. `0`
B. `log_(e)3`
C. `log_(e)4`
D. `log_(e)7`

Explanation

To solve the integral

\[ I = \int_{5}^{41} \frac{dx}{(f^{-1}(x))^5 + 5f^{-1}(x)} \]

where \( f(x) = x^5 + 5x - 1 \), we will follow these steps:

### Step 1: Find the values of \( f(1) \) and \( f(2) \)

First, we calculate \( f(1) \):

\[ f(1) = 1^5 + 5 \cdot 1 - 1 = 1 + 5 - 1 = 5 \]

Next, we calculate \( f(2) \):

\[ f(2) = 2^5 + 5 \cdot 2 - 1 = 32 + 10 - 1 = 41 \]

### Step 2: Identify the inverse function values

From the calculations above, we have:

\[ f^{-1}(5) = 1 \quad \text{and} \quad f^{-1}(41) = 2 \]

### Step 3: Substitute \( f^{-1}(x) \) with \( t \)

Let \( t = f^{-1}(x) \). Then, we have:

\[ x = f(t) = t^5 + 5t - 1 \]

### Step 4: Change the variable in the integral

The differential \( dx \) can be expressed in terms of \( dt \):

\[ dx = f'(t) dt \]

Calculating \( f'(t) \):

\[ f'(t) = 5t^4 + 5 \]

Thus, we can rewrite the integral:

\[ I = \int_{t=1}^{t=2} \frac{f'(t) dt}{t^5 + 5t} \]

### Step 5: Substitute \( f'(t) \) into the integral

Substituting \( f'(t) \):

\[ I = \int_{1}^{2} \frac{(5t^4 + 5) dt}{t^5 + 5t} \]

### Step 6: Factor out the common term

Factoring out \( 5 \):

\[ I = 5 \int_{1}^{2} \frac{(t^4 + 1) dt}{t^5 + 5t} \]

### Step 7: Change of variable for integration

Let \( u = t^5 + 5t \). Then, the differential \( du \) is:

\[ du = (5t^4 + 5) dt = 5(t^4 + 1) dt \]

This leads to:

\[ dt = \frac{du}{5(t^4 + 1)} \]

### Step 8: Change the limits of integration

When \( t = 1 \):

\[ u = 1^5 + 5 \cdot 1 = 6 \]

When \( t = 2 \):

\[ u = 2^5 + 5 \cdot 2 = 42 \]

### Step 9: Rewrite the integral in terms of \( u \)

Now the integral becomes:

\[ I = 5 \int_{6}^{42} \frac{du}{u} \]

### Step 10: Solve the integral

The integral of \( \frac{1}{u} \) is:

\[ \int \frac{du}{u} = \ln |u| \]

Thus, we have:

\[ I = 5 \left[ \ln |u| \right]_{6}^{42} = 5 (\ln 42 - \ln 6) \]

### Step 11: Simplify using properties of logarithms

Using the property of logarithms:

\[ I = 5 \ln \left( \frac{42}{6} \right) = 5 \ln 7 \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{5 \ln 7} \]