`int((1-x)^(2))/(x(1+x^(2)))dx=`

Options

A. `2logx-tan^(-1)x+c`
B. `logx-2tan^(-1)x+c`
C. `logx-tan^(-1)x+c`
D. `2logx-tan^(-1)x+c`

Answer

To solve the integral \( \int \frac{(1-x)^2}{x(1+x^2)} \, dx \), we can follow these steps:

### Step 1: Expand the numerator
First, we expand the numerator \( (1-x)^2 \):
\[ (1-x)^2 = 1 - 2x + x^2 \]
Thus, the integral becomes:
\[ \int \frac{1 - 2x + x^2}{x(1+x^2)} \, dx \]

### Step 2: Separate the integral
Next, we can separate the integral into three parts:
\[ \int \left( \frac{1}{x(1+x^2)} - \frac{2x}{x(1+x^2)} + \frac{x^2}{x(1+x^2)} \right) \, dx \]
This simplifies to:
\[ \int \left( \frac{1}{x(1+x^2)} - \frac{2}{1+x^2} + \frac{x}{1+x^2} \right) \, dx \]

### Step 3: Simplify each term
Now we can integrate each term separately:
1. For \( \int \frac{1}{x(1+x^2)} \, dx \), we can use partial fraction decomposition.
2. The second term \( -2 \int \frac{1}{1+x^2} \, dx \) is a standard integral.
3. The third term \( \int \frac{x}{1+x^2} \, dx \) can be solved using substitution.

### Step 4: Solve the first term using partial fractions
We can express:
\[ \frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx + C}{1+x^2} \]
Multiplying through by \( x(1+x^2) \) and solving for \( A, B, C \), we find:
\[ 1 = A(1+x^2) + (Bx + C)x \]
Setting up equations for coefficients gives us:
- For \( x^0 \): \( A = 1 \)
- For \( x^1 \): \( B = 0 \)
- For \( x^2 \): \( A + C = 0 \) implies \( C = -1 \)

Thus, we have:
\[ \frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{1}{1+x^2} \]

### Step 5: Integrate each term
Now we can integrate:
1. \( \int \frac{1}{x} \, dx = \ln |x| \)
2. \( -2 \int \frac{1}{1+x^2} \, dx = -2 \tan^{-1}(x) \)
3. For \( \int \frac{x}{1+x^2} \, dx \), we can use the substitution \( u = 1+x^2 \), \( du = 2x \, dx \), leading to:
\[ \frac{1}{2} \ln |1+x^2| \]

### Step 6: Combine results
Combining all the results, we get:
\[ \int \frac{(1-x)^2}{x(1+x^2)} \, dx = \ln |x| - \tan^{-1}(x) - \ln |1+x^2| + C \]

### Final Answer
Thus, the final answer is:
\[ \ln |x| - 2 \tan^{-1}(x) + \frac{1}{2} \ln |1+x^2| + C \]