If `x(x^(4)+1)phi(x)=1`, then `int_(1)^(2)phi(x)dx=`

Options

A. `1/4 "log" 32/17`
B. `1/2 "log" 32/17`
C. `1/4 "log"16/17`
D. `1/2 "log" 16/17`

Answer

To solve the problem, we need to evaluate the integral \( \int_{1}^{2} \phi(x) \, dx \) given the equation \( x(x^4 + 1) \phi(x) = 1 \). Let's break it down step by step.

### Step 1: Express \( \phi(x) \)
From the given equation, we can isolate \( \phi(x) \):
\[ \phi(x) = \frac{1}{x(x^4 + 1)} \]

### Step 2: Set up the integral
Now, we can substitute \( \phi(x) \) into the integral:
\[ \int_{1}^{2} \phi(x) \, dx = \int_{1}^{2} \frac{1}{x(x^4 + 1)} \, dx \]

### Step 3: Simplify the integrand
We can rewrite the integrand:
\[ \frac{1}{x(x^4 + 1)} = \frac{1}{x^5 + x} \]
This will help us in the next steps.

### Step 4: Use substitution
To evaluate the integral, we can use the substitution \( t = x^4 + 1 \). Then, we differentiate:
\[ dt = 4x^3 \, dx \implies dx = \frac{dt}{4x^3} \]
We also need to change the limits of integration. When \( x = 1 \):
\[ t = 1^4 + 1 = 2 \]
When \( x = 2 \):
\[ t = 2^4 + 1 = 17 \]

### Step 5: Rewrite the integral in terms of \( t \)
Now we can rewrite the integral:
\[ \int_{1}^{2} \frac{1}{x(x^4 + 1)} \, dx = \int_{2}^{17} \frac{1}{x(t)} \cdot \frac{dt}{4x^3} \]
Since \( x^4 = t - 1 \), we can express \( x \) in terms of \( t \):
\[ x = (t - 1)^{1/4} \]
Thus, the integral becomes:
\[ \int_{2}^{17} \frac{1}{(t - 1)^{1/4}((t - 1)^{1/4})^4 + 1} \cdot \frac{dt}{4(t - 1)^{3/4}} \]
This simplifies to:
\[ \int_{2}^{17} \frac{1}{4(t - 1)^{1}} \, dt \]

### Step 6: Evaluate the integral
Now we can evaluate the integral:
\[ \frac{1}{4} \int_{2}^{17} \frac{1}{t - 1} \, dt \]
The integral of \( \frac{1}{t - 1} \) is \( \ln |t - 1| \):
\[ = \frac{1}{4} \left[ \ln |t - 1| \right]_{2}^{17} \]
Calculating the limits:
\[ = \frac{1}{4} \left( \ln(16) - \ln(1) \right) = \frac{1}{4} \ln(16) \]
Since \( \ln(1) = 0 \):
\[ = \frac{1}{4} \cdot 4 \ln(2) = \ln(2) \]

### Final Result
Thus, the value of the integral \( \int_{1}^{2} \phi(x) \, dx \) is:
\[ \boxed{\ln(2)} \]