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 step by step, we will follow the instructions given in the question and the video transcript.

### Step 1: Define the Functions
Given:
- \( f(x) = 2x + 1 \)
- \( g(x) = \int \frac{f(x)}{x^2 (x + 1)^2} \, dx \)

### Step 2: Rewrite the Integral
Substituting \( f(x) \) into the integral for \( g(x) \):
\[ g(x) = \int \frac{2x + 1}{x^2 (x + 1)^2} \, dx \]

### Step 3: Simplify the Integral
We can split the fraction:
\[ g(x) = \int \left( \frac{2x + 1}{x^2 (x + 1)^2} \right) \, dx = \int \left( \frac{2x}{x^2 (x + 1)^2} + \frac{1}{x^2 (x + 1)^2} \right) \, dx \]

### Step 4: Rewrite the Numerator
We can rewrite the numerator \( 2x + 1 \) as follows:
\[ 2x + 1 = (x^2 + 2x + 1) - x^2 = (x + 1)^2 - x^2 \]
Thus, we have:
\[ g(x) = \int \left( \frac{(x + 1)^2 - x^2}{x^2 (x + 1)^2} \right) \, dx \]
This simplifies to:
\[ g(x) = \int \left( \frac{1}{x^2} - \frac{1}{(x + 1)^2} \right) \, dx \]

### Step 5: Integrate
Now we can integrate each term:
\[ g(x) = \int \frac{1}{x^2} \, dx - \int \frac{1}{(x + 1)^2} \, dx \]
The integrals yield:
\[ g(x) = -\frac{1}{x} + \frac{1}{x + 1} + C \]

### Step 6: Apply the Given Condition
We are given that:
\[ 6g(2) + 1 = 0 \]
Calculating \( g(2) \):
\[ g(2) = -\frac{1}{2} + \frac{1}{3} + C = -\frac{3}{6} + \frac{2}{6} + C = -\frac{1}{6} + C \]
Substituting into the condition:
\[ 6\left(-\frac{1}{6} + C\right) + 1 = 0 \]
This simplifies to:
\[ -1 + 6C + 1 = 0 \implies 6C = 0 \implies C = 0 \]

### Step 7: Final Form of \( g(x) \)
Thus, we have:
\[ g(x) = -\frac{1}{x} + \frac{1}{x + 1} \]

### Step 8: Calculate \( g\left(-\frac{1}{2}\right) \)
Now we need to find \( g\left(-\frac{1}{2}\right) \):
\[ g\left(-\frac{1}{2}\right) = -\frac{1}{-\frac{1}{2}} + \frac{1}{-\frac{1}{2} + 1} \]
Calculating each term:
\[ g\left(-\frac{1}{2}\right) = 2 + \frac{1}{\frac{1}{2}} = 2 + 2 = 4 \]

### Final Answer
Thus, \( g\left(-\frac{1}{2}\right) = 4 \).

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