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 \( f(x) = \frac{x}{x+1} \) given that \( x \) is in \( \mathbb{R} \setminus \{-1, 0\} \).

### Step-by-Step Solution:

1. **Start with the given integral**:
\[ \int \frac{dx}{x^2 + x} = \ln |f(x)| + C \]

2. **Simplify the integrand**:
We can factor the denominator:
\[ x^2 + x = x(x + 1) \]
Thus, we rewrite the integral:
\[ \int \frac{dx}{x(x + 1)} \]

3. **Use partial fraction decomposition**:
We can express the integrand as:
\[ \frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \]
Multiplying through by the denominator \( x(x + 1) \) gives:
\[ 1 = A(x + 1) + Bx \]
Setting \( x = 0 \) gives \( A = 1 \). Setting \( x = -1 \) gives \( B = -1 \). Therefore:
\[ \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1} \]

4. **Integrate each term**:
Now we can integrate:
\[ \int \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx = \ln |x| - \ln |x + 1| + C \]
This simplifies to:
\[ \ln \left| \frac{x}{x + 1} \right| + C \]

5. **Set the expression equal to \( \ln |f(x)| + C \)**:
From the original equation, we have:
\[ \ln \left| \frac{x}{x + 1} \right| = \ln |f(x)| \]
Thus, we can conclude:
\[ f(x) = \frac{x}{x + 1} \]

6. **Find the range of \( f(x) \)**:
To find the range of \( f(x) = \frac{x}{x + 1} \):
- As \( x \to -1 \), \( f(x) \to -\infty \).
- As \( x \to 0 \), \( f(x) \to 0 \).
- As \( x \to \infty \), \( f(x) \to 1 \).
- As \( x \to -\infty \), \( f(x) \to 1 \).

The function is continuous and takes all values except at \( y = 0 \) and \( y = 1 \).

7. **Conclusion**:
Therefore, the range of \( y = f(x) \) is:
\[ y \in \mathbb{R} \setminus \{0, 1\} \]