If `int((2x+1)dx)/(x^(4)+2x^(3)+x^(2)-1)=Aln|(x^(2)+x+1)/(x^(2)+x-1)|+C`, then

Options

A. A=1
B. A=1/2
C. A=-1/2
D. None of these

Answer

To solve the integral

\[ I = \int \frac{2x + 1}{x^4 + 2x^3 + x^2 - 1} \, dx, \]

we will follow a systematic approach.

### Step 1: Factor the Denominator
First, we need to simplify the denominator \(x^4 + 2x^3 + x^2 - 1\). We can factor out \(x^2\) from the first three terms:

\[ x^4 + 2x^3 + x^2 - 1 = x^2(x^2 + 2x + 1) - 1 = x^2(x + 1)^2 - 1. \]

Now we can recognize this as a difference of squares:

\[ x^2(x + 1)^2 - 1^2 = (x(x + 1) - 1)(x(x + 1) + 1). \]

Thus, we can rewrite the denominator as:

\[ (x(x + 1) - 1)(x(x + 1) + 1). \]

### Step 2: Substitute Variables
Next, we will use the substitution \(t = x^2 + x\). Then, we differentiate:

\[ \frac{dt}{dx} = 2x + 1 \implies dt = (2x + 1)dx. \]

This allows us to replace \(2x + 1 \, dx\) in the integral:

\[ I = \int \frac{dt}{(t + 1)(t - 1)}. \]

### Step 3: Partial Fraction Decomposition
Now we will perform partial fraction decomposition on the integrand:

\[ \frac{1}{(t + 1)(t - 1)} = \frac{A}{t + 1} + \frac{B}{t - 1}. \]

Multiplying through by the denominator \((t + 1)(t - 1)\) gives:

\[ 1 = A(t - 1) + B(t + 1). \]

Expanding and rearranging, we get:

\[ 1 = (A + B)t + (B - A). \]

Setting coefficients equal, we have:

1. \(A + B = 0\)
2. \(B - A = 1\)

From the first equation, \(B = -A\). Substituting into the second equation:

\[ -A - A = 1 \implies -2A = 1 \implies A = -\frac{1}{2}, \quad B = \frac{1}{2}. \]

Thus, we can rewrite the integral:

\[ I = \int \left(-\frac{1}{2(t + 1)} + \frac{1}{2(t - 1)}\right) dt. \]

### Step 4: Integrate
Now we can integrate term by term:

\[ I = -\frac{1}{2} \int \frac{1}{t + 1} dt + \frac{1}{2} \int \frac{1}{t - 1} dt. \]

This gives us:

\[ I = -\frac{1}{2} \ln |t + 1| + \frac{1}{2} \ln |t - 1| + C. \]

### Step 5: Combine Logarithms
Using properties of logarithms, we can combine these:

\[ I = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C. \]

### Step 6: Substitute Back
Substituting back \(t = x^2 + x\):

\[ I = \frac{1}{2} \ln \left| \frac{x^2 + x - 1}{x^2 + x + 1} \right| + C. \]

### Conclusion
Comparing with the given expression \(A \ln \left| \frac{x^2 + x + 1}{x^2 + x - 1} \right| + C\), we find:

\[ A = \frac{1}{2}. \]

Thus, the final answer is:

\[ A = \frac{1}{2}. \]