If `int (x^4 + 1)/(x(x^2+1)^2)\ dx = A ln |x| +B/(1+x^2)+C,` then `A+B` equals to :

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To solve the integral

\[ \int \frac{x^4 + 1}{x(x^2 + 1)^2} \, dx, \]

we will simplify the integrand and then integrate term by term.

### Step 1: Simplifying the Integrand
We start with the expression:

\[ \frac{x^4 + 1}{x(x^2 + 1)^2}. \]

We can rewrite \(x^4 + 1\) by adding and subtracting \(2x^2\):

\[ x^4 + 1 = x^4 + 2x^2 - 2x^2 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - 2x^2. \]

Thus, we can rewrite the integrand as:

\[ \frac{(x^2 + 1)^2 - 2x^2}{x(x^2 + 1)^2} = \frac{(x^2 + 1)^2}{x(x^2 + 1)^2} - \frac{2x^2}{x(x^2 + 1)^2}. \]

This simplifies to:

\[ \frac{1}{x} - \frac{2x}{(x^2 + 1)^2}. \]

### Step 2: Splitting the Integral
Now we can split the integral into two parts:

\[ \int \left( \frac{1}{x} - \frac{2x}{(x^2 + 1)^2} \right) \, dx = \int \frac{1}{x} \, dx - 2 \int \frac{x}{(x^2 + 1)^2} \, dx. \]

### Step 3: Integrating the First Part
The first integral is straightforward:

\[ \int \frac{1}{x} \, dx = \ln |x|. \]

### Step 4: Integrating the Second Part
For the second integral, we use the substitution \(u = x^2 + 1\), which gives \(du = 2x \, dx\) or \(dx = \frac{du}{2x}\). Thus,

\[ \int \frac{x}{(x^2 + 1)^2} \, dx = \int \frac{x}{u^2} \cdot \frac{du}{2x} = \frac{1}{2} \int \frac{1}{u^2} \, du. \]

Integrating \(\frac{1}{u^2}\) gives:

\[ \frac{1}{2} \left( -\frac{1}{u} \right) = -\frac{1}{2(x^2 + 1)}. \]

### Step 5: Combining the Results
Putting it all together, we have:

\[ \int \frac{x^4 + 1}{x(x^2 + 1)^2} \, dx = \ln |x| - 2 \left( -\frac{1}{2(x^2 + 1)} \right) + C = \ln |x| + \frac{1}{x^2 + 1} + C. \]

### Step 6: Identifying Constants A, B, and C
From the expression:

\[ \int \frac{x^4 + 1}{x(x^2 + 1)^2} \, dx = A \ln |x| + \frac{B}{1 + x^2} + C, \]

we can identify:

- \(A = 1\)
- \(B = 1\)

### Final Step: Calculating A + B
Thus,

\[ A + B = 1 + 1 = 2. \]

### Conclusion
The final answer is:

\[ \boxed{2}. \]