The value of `int_(1)^(2) (dx)/(x(1+x^(4))` is

Options

A. `(1)/(4)log.(17)/(32)`
B. `(1)/(4)log.(32)/(17)`
C. `log.(17)/(2)`
D. `(1)/(4)log.(17)/(2)`

Explanation

To solve the integral \( I = \int_{1}^{2} \frac{dx}{x(1+x^4)} \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int_{1}^{2} \frac{dx}{x(1+x^4)} \]

### Step 2: Substitute \( x^4 \)
Let's make the substitution \( t = x^4 \). Then, we differentiate to find \( dt \):
\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]
Also, we need to express \( x \) in terms of \( t \):
\[ x = t^{1/4} \quad \Rightarrow \quad x^3 = (t^{1/4})^3 = t^{3/4} \]

### Step 3: Change the limits of integration
When \( x = 1 \):
\[ t = 1^4 = 1 \]
When \( x = 2 \):
\[ t = 2^4 = 16 \]
Thus, the limits change from \( x = 1 \) to \( x = 2 \) to \( t = 1 \) to \( t = 16 \).

### Step 4: Substitute into the integral
Now substituting \( dx \) and changing the limits:
\[ I = \int_{1}^{16} \frac{1}{t^{1/4}(1+t)} \cdot \frac{dt}{4t^{3/4}} = \frac{1}{4} \int_{1}^{16} \frac{dt}{t(1+t)} \]

### Step 5: Partial fraction decomposition
We can decompose \( \frac{1}{t(1+t)} \) into partial fractions:
\[ \frac{1}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t} \]
Multiplying through by \( t(1+t) \):
\[ 1 = A(1+t) + Bt \]
Setting \( t = 0 \):
\[ 1 = A \quad \Rightarrow \quad A = 1 \]
Setting \( t = -1 \):
\[ 1 = B(-1) \quad \Rightarrow \quad B = -1 \]
Thus, we have:
\[ \frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t} \]

### Step 6: Substitute back into the integral
Now substituting back into the integral:
\[ I = \frac{1}{4} \int_{1}^{16} \left( \frac{1}{t} - \frac{1}{1+t} \right) dt \]

### Step 7: Integrate
Now we can integrate:
\[ \int \frac{1}{t} dt = \log t \quad \text{and} \quad \int \frac{1}{1+t} dt = \log(1+t) \]
Thus,
\[ I = \frac{1}{4} \left[ \log t - \log(1+t) \right]_{1}^{16} \]

### Step 8: Evaluate the limits
Now we evaluate at the limits:
\[ I = \frac{1}{4} \left[ \left( \log 16 - \log 17 \right) - \left( \log 1 - \log 2 \right) \right] \]
Since \( \log 1 = 0 \):
\[ I = \frac{1}{4} \left[ \log 16 - \log 17 + \log 2 \right] \]
Using the properties of logarithms:
\[ I = \frac{1}{4} \left[ \log \frac{16 \cdot 2}{17} \right] = \frac{1}{4} \left[ \log \frac{32}{17} \right] \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{1}{4} \log \frac{32}{17} \]