Let `f(x)=2x+1 and g(x)=int(f(x))/(x^(2)(x+1)^(2))dx`. If `6g(2)+1=0` then `g(-(1)/(2))` is equal to

Options

A. 4
B. `-4`
C. 3
D. 2

Answer

To solve the problem, we will follow these steps:

1. **Define the functions**: We are given \( f(x) = 2x + 1 \) and \( g(x) = \int \frac{f(x)}{x^2 (x+1)^2} \, dx \).

2. **Rewrite the integral**: Substitute \( f(x) \) into the integral:
\[ g(x) = \int \frac{2x + 1}{x^2 (x+1)^2} \, dx \]

3. **Simplify the integrand**: We can separate the integrand:
\[ g(x) = \int \left( \frac{2x}{x^2 (x+1)^2} + \frac{1}{x^2 (x+1)^2} \right) \, dx \]

4. **Break down the fractions**: We can simplify \( \frac{2x}{x^2 (x+1)^2} \) and \( \frac{1}{x^2 (x+1)^2} \):
\[ \frac{2x}{x^2 (x+1)^2} = \frac{2}{x(x+1)^2} \]
\[ \frac{1}{x^2 (x+1)^2} = \frac{1}{x^2} - \frac{1}{(x+1)^2} \]

5. **Combine the fractions**: Thus, we can write:
\[ g(x) = \int \left( \frac{2}{x(x+1)^2} + \frac{1}{x^2} - \frac{1}{(x+1)^2} \right) \, dx \]

6. **Integrate each term**:
- The integral of \( \frac{2}{x(x+1)^2} \) can be solved using partial fractions.
- The integral of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \).
- The integral of \( \frac{1}{(x+1)^2} \) is \( -\frac{1}{x+1} \).

7. **Combine the results**:
\[ g(x) = -\frac{1}{x} + \frac{1}{x+1} + C \]

8. **Use the condition**: We have the condition \( 6g(2) + 1 = 0 \):
\[ g(2) = -\frac{1}{2} + \frac{1}{3} + C \]
\[ g(2) = -\frac{3}{6} + \frac{2}{6} + C = -\frac{1}{6} + C \]
Setting \( 6(-\frac{1}{6} + C) + 1 = 0 \):
\[ -1 + 6C + 1 = 0 \implies 6C = 0 \implies C = 0 \]

9. **Final form of \( g(x) \)**:
\[ g(x) = -\frac{1}{x} + \frac{1}{x+1} \]

10. **Calculate \( g\left(-\frac{1}{2}\right) \)**:
\[ g\left(-\frac{1}{2}\right) = -\frac{1}{-\frac{1}{2}} + \frac{1}{-\frac{1}{2} + 1} = 2 + \frac{1}{\frac{1}{2}} = 2 + 2 = 4 \]

Thus, the value of \( g\left(-\frac{1}{2}\right) \) is \( 4 \).