`int1/x^(2)(2x+1)^(3)dx=`

Options

A. `4x^(2)+12x+6logx-1/x+c`
B. `4x^(2)+12x-6logx-2/x+c`
C. `2x^(2)+8x+3logx-2/x+c`
D. `8x^(2)+6x+6logx+2/x+c`

Answer

To solve the integral \( \int \frac{1}{x^2 (2x + 1)^3} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{x^2 (2x + 1)^3} \, dx \]

### Step 2: Expand the Denominator
We can expand \( (2x + 1)^3 \) using the binomial theorem:
\[ (2x + 1)^3 = 8x^3 + 12x^2 + 6x + 1 \]
Thus, we can rewrite our integral as:
\[ \int \frac{1}{x^2 (8x^3 + 12x^2 + 6x + 1)} \, dx \]

### Step 3: Simplify the Expression
Now we can separate the terms in the denominator:
\[ \int \left( \frac{8x^3}{x^2} + \frac{12x^2}{x^2} + \frac{6x}{x^2} + \frac{1}{x^2} \right)^{-1} \, dx = \int \left( 8x + 12 + \frac{6}{x} + \frac{1}{x^2} \right)^{-1} \, dx \]

### Step 4: Rewrite the Integral
We can rewrite the integral as:
\[ \int \left( 8x + 12 + \frac{6}{x} + \frac{1}{x^2} \right)^{-1} \, dx \]

### Step 5: Integrate Each Term
Now we can integrate each term separately:
\[ \int \left( 8x + 12 + \frac{6}{x} + \frac{1}{x^2} \right)^{-1} \, dx = \int 8x \, dx + \int 12 \, dx + \int \frac{6}{x} \, dx + \int \frac{1}{x^2} \, dx \]

Calculating each integral:
1. \( \int 8x \, dx = 4x^2 + C_1 \)
2. \( \int 12 \, dx = 12x + C_2 \)
3. \( \int \frac{6}{x} \, dx = 6 \ln |x| + C_3 \)
4. \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C_4 \)

### Step 6: Combine the Results
Combining all the results, we have:
\[ 4x^2 + 12x + 6 \ln |x| - \frac{1}{x} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{x^2 (2x + 1)^3} \, dx = 4x^2 - \frac{1}{x} + 12x + 6 \ln |x| + C \]