`int(1)/(x(x^(4)-1))dx` is equal to

Options

A. `(1)/(4)log|(x^(4))/(x^(4)-1)|+C`
B. `(1)/(4)log|(x^(4)-1)/(x^(4))|+C`
C. `log|(x^(4)-1)/(x^(4))|+C`
D. `log|(x^(4))/(x^(4)-1)|+C`

Answer

To solve the integral \( I = \int \frac{1}{x(x^4 - 1)} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{x(x^4 - 1)} \, dx \]
To simplify the expression, we can multiply and divide by \( x^3 \):
\[ I = \int \frac{x^3}{x^4(x^4 - 1)} \, dx \]

### Step 2: Substitution
Let \( t = x^4 \). Then, differentiate both sides:
\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]
Substituting \( x^3 \, dx \) into the integral:
\[ I = \int \frac{1}{t(t - 1)} \cdot \frac{dt}{4} = \frac{1}{4} \int \frac{1}{t(t - 1)} \, dt \]

### Step 3: Partial Fraction Decomposition
Now we need to decompose \( \frac{1}{t(t - 1)} \) into partial fractions:
\[ \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \]
Multiplying through by the denominator \( t(t - 1) \):
\[ 1 = A(t - 1) + Bt \]
Expanding and rearranging gives:
\[ 1 = At - A + Bt \quad \Rightarrow \quad 1 = (A + B)t - A \]
From this, we can equate coefficients:
1. \( A + B = 0 \)
2. \( -A = 1 \)

Solving these equations, we find:
- From \( -A = 1 \), we have \( A = -1 \).
- Substituting \( A \) into \( A + B = 0 \) gives \( -1 + B = 0 \) or \( B = 1 \).

Thus, we have:
\[ \frac{1}{t(t - 1)} = \frac{-1}{t} + \frac{1}{t - 1} \]

### Step 4: Integrate
Substituting back into the integral:
\[ I = \frac{1}{4} \left( \int \frac{-1}{t} \, dt + \int \frac{1}{t - 1} \, dt \right) \]
Calculating the integrals:
\[ I = \frac{1}{4} \left( -\ln |t| + \ln |t - 1| \right) + C \]
This simplifies to:
\[ I = \frac{1}{4} \ln \left| \frac{t - 1}{t} \right| + C \]

### Step 5: Substitute Back
Now, substituting back \( t = x^4 \):
\[ I = \frac{1}{4} \ln \left| \frac{x^4 - 1}{x^4} \right| + C \]
This can be simplified further:
\[ I = \frac{1}{4} \ln \left| \frac{x^4 - 1}{x^4} \right| + C = \frac{1}{4} \left( \ln |x^4 - 1| - \ln |x^4| \right) + C \]
Using properties of logarithms:
\[ I = \frac{1}{4} \ln \left| \frac{x^4 - 1}{x^4} \right| + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{1}{4} \ln \left| \frac{x^4 - 1}{x^4} \right| + C \]