`int(dx)/(x-x^3)=`

Options

A. `(1)/(2)log|(1-x^2)/(x^2)|+c`
B. `log|(1-x)/(x(1+x))|+c`
C. `log|x-x^3|+c`
D. `(1)/(2)log|(x^2)/(1-x^2)|+c`

Answer

To solve the integral \(\int \frac{dx}{x - x^3}\), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Factor the Denominator
We start with the expression in the denominator:
\[ x - x^3 = x(1 - x^2) = x(1 - x)(1 + x) \]
Thus, we can rewrite our integral as:
\[ \int \frac{dx}{x(1 - x)(1 + x)} \]

### Step 2: Set Up Partial Fractions
Next, we express the integrand as a sum of partial fractions:
\[ \frac{1}{x(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x} \]
where \(A\), \(B\), and \(C\) are constants we need to determine.

### Step 3: Clear the Denominator
Multiply both sides by the denominator \(x(1 - x)(1 + x)\):
\[ 1 = A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x) \]

### Step 4: Expand and Collect Terms
Expanding the right-hand side gives:
\[ 1 = A(1 - x^2) + Bx + Bx^2 + Cx - Cx^2 \]
Combining like terms:
\[ 1 = A - Ax^2 + (B + C)x + (B - C)x^2 \]

### Step 5: Set Up Equations
Now, we equate coefficients from both sides:
1. Constant term: \(A = 1\)
2. Coefficient of \(x\): \(B + C = 0\)
3. Coefficient of \(x^2\): \(-A + B - C = 0\)

### Step 6: Solve the System of Equations
From \(A = 1\), we substitute into the other equations:
- From \(B + C = 0\), we have \(C = -B\).
- Substituting into \(-1 + B - (-B) = 0\) gives us:
\[ -1 + 2B = 0 \implies B = \frac{1}{2} \]
Thus, \(C = -\frac{1}{2}\).

### Step 7: Write the Partial Fraction Decomposition
Now we have:
\[ \frac{1}{x(1 - x)(1 + x)} = \frac{1}{x} + \frac{1/2}{1 - x} - \frac{1/2}{1 + x} \]

### Step 8: Integrate Each Term
Now we can integrate each term separately:
\[ \int \left( \frac{1}{x} + \frac{1/2}{1 - x} - \frac{1/2}{1 + x} \right) dx \]
This gives:
\[ \int \frac{1}{x} \, dx + \frac{1}{2} \int \frac{1}{1 - x} \, dx - \frac{1}{2} \int \frac{1}{1 + x} \, dx \]

### Step 9: Perform the Integrations
Calculating these integrals:
1. \(\int \frac{1}{x} \, dx = \ln |x|\)
2. \(\int \frac{1}{1 - x} \, dx = -\ln |1 - x|\)
3. \(\int \frac{1}{1 + x} \, dx = \ln |1 + x|\)

Putting it all together:
\[ \ln |x| - \frac{1}{2} \ln |1 - x| - \frac{1}{2} \ln |1 + x| + C \]

### Step 10: Combine the Logarithms
Using properties of logarithms:
\[ = \ln |x| - \frac{1}{2} \ln |(1 - x)(1 + x)| + C \]
This can be rewritten as:
\[ = \ln \left( \frac{|x|}{\sqrt{(1 - x)(1 + x)}} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{x - x^3} = \ln \left( \frac{|x|}{\sqrt{1 - x^2}} \right) + C \]