If `int(dx)/(x^(2)+x)=ln|f(x)|+C` (where C is the constant of integration), then the range of `y=f(x), AA x in R-{-1, 0}` is

Options

A. `R-{1}`
B. `R-{0}`
C. `R-{0, 1}`
D. `R-{0, -1}`

Answer

To solve the problem, we need to find the range of the function \( f(x) \) given the integral equation:

\[ \int \frac{dx}{x^2 + x} = \ln |f(x)| + C \]

### Step 1: Simplify the Integral

First, we simplify the integrand:

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

Next, we can use partial fraction decomposition:

\[ \frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \]

Multiplying through by \( x(x + 1) \) gives:

\[ 1 = A(x + 1) + Bx \]

Setting \( x = 0 \):

\[ 1 = A(0 + 1) \implies A = 1 \]

Setting \( x = -1 \):

\[ 1 = B(-1) \implies B = -1 \]

Thus, we have:

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

### Step 2: Integrate

Now we can integrate:

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

Using the properties of logarithms, we can combine these:

\[ \ln \left| \frac{x}{x + 1} \right| + C \]

### Step 3: Set Equal to \( \ln |f(x)| \)

According to the problem, we have:

\[ \ln \left| \frac{x}{x + 1} \right| + C = \ln |f(x)| \]

This implies:

\[ |f(x)| = e^C \cdot \left| \frac{x}{x + 1} \right| \]

Let \( k = e^C \), then:

\[ |f(x)| = k \cdot \left| \frac{x}{x + 1} \right| \]

### Step 4: Analyze the Function

Now we need to analyze the function \( \frac{x}{x + 1} \):

1. The function is undefined at \( x = -1 \) and \( x = 0 \).
2. As \( x \to -1^+ \), \( \frac{x}{x + 1} \to -\infty \).
3. As \( x \to 0^- \), \( \frac{x}{x + 1} \to 0^- \).
4. As \( x \to 0^+ \), \( \frac{x}{x + 1} \to 0^+ \).
5. As \( x \to \infty \), \( \frac{x}{x + 1} \to 1 \).
6. As \( x \to -\infty \), \( \frac{x}{x + 1} \to 1 \).

### Step 5: Determine the Range of \( f(x) \)

From the analysis, we can conclude:

- The function \( \frac{x}{x + 1} \) takes values in \( (-\infty, 0) \) as \( x \) approaches \( -1 \) and \( 0 \).
- The function approaches \( 1 \) but never reaches it as \( x \to \infty \).

Thus, the range of \( |f(x)| \) is:

\[ (0, \infty) \]

Since \( f(x) \) can be either positive or negative, the complete range of \( f(x) \) is:

\[ (-\infty, 0) \cup (0, \infty) \]

### Final Answer

The range of \( y = f(x) \) for \( x \in \mathbb{R} \setminus \{-1, 0\} \) is:

\[ (-\infty, 0) \cup (0, \infty) \]