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

Options

A. `logx+c`
B. `(logx)(tan^(-1)x)+c`
C. `2logx+tan^(-1)x+c`
D. `logx+2tan^(-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: Split the integral
Next, we can separate the integral into three parts:
\[ \int \frac{1}{x(1+x^2)} \, dx + \int \frac{2x}{x(1+x^2)} \, dx + \int \frac{x^2}{x(1+x^2)} \, dx \]
This simplifies to:
\[ \int \frac{1}{x(1+x^2)} \, dx + \int \frac{2}{1+x^2} \, dx + \int \frac{x}{1+x^2} \, dx \]

### Step 3: Solve each integral separately
1. **First Integral**: \( \int \frac{1}{x(1+x^2)} \, dx \)

We can use partial fraction decomposition:
\[ \frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx + C}{1+x^2} \]
Multiplying through by the denominator \( x(1+x^2) \) and solving for \( A, B, C \) gives:
\[ 1 = A(1+x^2) + (Bx + C)x \]
Setting \( x = 0 \) gives \( A = 1 \). By substituting other values or comparing coefficients, we find \( B = 0 \) and \( C = -1 \). Thus:
\[ \frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{1}{1+x^2} \]
Therefore:
\[ \int \frac{1}{x(1+x^2)} \, dx = \ln |x| - \tan^{-1}(x) \]

2. **Second Integral**: \( \int \frac{2}{1+x^2} \, dx \)
This integral is straightforward:
\[ \int \frac{2}{1+x^2} \, dx = 2 \tan^{-1}(x) \]

3. **Third Integral**: \( \int \frac{x}{1+x^2} \, dx \)
We can use the substitution \( u = 1 + x^2 \), \( du = 2x \, dx \):
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln |1+x^2| \]

### Step 4: Combine the results
Now, we can combine all the results:
\[ \int \frac{(1+x)^2}{x(1+x^2)} \, dx = \left( \ln |x| - \tan^{-1}(x) \right) + 2 \tan^{-1}(x) + \frac{1}{2} \ln |1+x^2| + C \]
This simplifies to:
\[ \int \frac{(1+x)^2}{x(1+x^2)} \, dx = \ln |x| + \tan^{-1}(x) + \frac{1}{2} \ln |1+x^2| + C \]

### Final Answer
\[ \int \frac{(1+x)^2}{x(1+x^2)} \, dx = \ln |x| + \tan^{-1}(x) + \frac{1}{2} \ln |1+x^2| + C \]