If `int(1)/(x^(3)+x^(4))dx=(A)/(x^(2))+(B)/(x)+log|(x)/(x+1)|+C` , then

Options

A. `A=(1)/(2),B=1`
B. `A=1,B=-(1)/(2)`
C. `A=-(1)/(2),B=1`
D. none of these

Answer

To solve the integral \( \int \frac{1}{x^3 + x^4} \, dx \) and express it in the form \( \frac{A}{x^2} + \frac{B}{x} + \log\left|\frac{x}{x+1}\right| + C \), we will follow these steps:

### Step 1: Simplify the integrand
First, we can factor the denominator:
\[ x^3 + x^4 = x^3(1 + x) \]
Thus, we can rewrite the integral as:
\[ \int \frac{1}{x^3(1+x)} \, dx \]

### Step 2: Use Partial Fraction Decomposition
We will express \( \frac{1}{x^3(1+x)} \) using partial fractions:
\[ \frac{1}{x^3(1+x)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{1+x} \]
To find the coefficients \( A, B, C, D \), we multiply both sides by the denominator \( x^3(1+x) \):
\[ 1 = A x^2(1+x) + B x(1+x) + C(1+x) + D x^3 \]

### Step 3: Expand and Collect Terms
Expanding the right-hand side:
\[ 1 = A x^2 + A x^3 + B x + B x^2 + C + C x + D x^3 \]
Combining like terms gives:
\[ 1 = (A + D)x^3 + (A + B)x^2 + (B + C)x + C \]

### Step 4: Set Up the System of Equations
Now, we can set up the equations based on the coefficients:
1. \( A + D = 0 \) (coefficient of \( x^3 \))
2. \( A + B = 0 \) (coefficient of \( x^2 \))
3. \( B + C = 0 \) (coefficient of \( x \))
4. \( C = 1 \) (constant term)

### Step 5: Solve the System
From equation 4, we have \( C = 1 \).
Substituting \( C \) into equation 3:
\[ B + 1 = 0 \implies B = -1 \]
Substituting \( B \) into equation 2:
\[ A - 1 = 0 \implies A = 1 \]
Substituting \( A \) into equation 1:
\[ 1 + D = 0 \implies D = -1 \]

### Step 6: Write the Partial Fraction Decomposition
Now we have:
\[ \frac{1}{x^3(1+x)} = \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} - \frac{1}{1+x} \]

### Step 7: Integrate Each Term
Now we can integrate each term:
\[ \int \left( \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} - \frac{1}{1+x} \right) dx \]
This gives:
\[ \int \frac{1}{x} \, dx - \int \frac{1}{x^2} \, dx + \int \frac{1}{x^3} \, dx - \int \frac{1}{1+x} \, dx \]
Calculating each integral:
1. \( \int \frac{1}{x} \, dx = \log|x| \)
2. \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} \)
3. \( \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} \)
4. \( \int \frac{1}{1+x} \, dx = \log|1+x| \)

### Step 8: Combine the Results
Combining these results, we have:
\[ \log|x| - \left(-\frac{1}{x}\right) - \frac{1}{2x^2} - \log|1+x| + C \]
This simplifies to:
\[ \log\left|\frac{x}{1+x}\right| + \frac{1}{x} - \frac{1}{2x^2} + C \]

### Final Form
Thus, we can express the integral as:
\[ \int \frac{1}{x^3 + x^4} \, dx = \frac{1}{x} - \frac{1}{2x^2} + \log\left|\frac{x}{1+x}\right| + C \]
Comparing with the given form, we find:
- \( A = -\frac{1}{2} \)
- \( B = 1 \)