`int((x+1)^(2)dx)/(x(x^(2)+1))` is equal to

Options

A. `log_(e)x+c`
B. `log_(e)x+2tan^(-1)x+c`
C. `log_(e).(1)/(x^(2)+1)+c`
D. `log_(e){x(x^(2)+1)}+c`

Explanation

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

### Step 1: Expand the numerator
First, we expand \( (x+1)^2 \):
\[ (x+1)^2 = x^2 + 2x + 1 \]
So, we can rewrite the integral as:
\[ \int \frac{x^2 + 2x + 1}{x(x^2 + 1)} \, dx \]

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

### Step 3: Simplify each term
Now we can integrate each term separately:
1. For \( \int \frac{x}{x^2 + 1} \, dx \), we can use the substitution \( u = x^2 + 1 \), \( du = 2x \, dx \), hence \( \frac{1}{2} du = x \, dx \):
\[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln |x^2 + 1| + C_1 \]

2. For \( \int \frac{2}{x^2 + 1} \, dx \):
\[ \int \frac{2}{x^2 + 1} \, dx = 2 \tan^{-1}(x) + C_2 \]

3. For \( \int \frac{1}{x(x^2 + 1)} \, dx \), we can use partial fraction decomposition:
\[ \frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \]
Multiplying through by the denominator \( x(x^2 + 1) \) and equating coefficients, we find:
\[ 1 = A(x^2 + 1) + (Bx + C)x \]
Solving gives \( A = 1 \), \( B = 0 \), and \( C = -1 \). Thus:
\[ \int \frac{1}{x(x^2 + 1)} \, dx = \int \left( \frac{1}{x} - \frac{1}{x^2 + 1} \right) \, dx = \ln |x| - \tan^{-1}(x) + C_3 \]

### Step 4: Combine results
Now we combine all the integrals:
\[ \int \frac{(x+1)^2}{x(x^2 + 1)} \, dx = \frac{1}{2} \ln |x^2 + 1| + 2 \tan^{-1}(x) + \ln |x| - \tan^{-1}(x) + C \]
Simplifying, we have:
\[ = \frac{1}{2} \ln |x^2 + 1| + \tan^{-1}(x) + \ln |x| + C \]

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