If `int(f(x))/(x^(3)-1)dx`, where `f(x)` is a polynomial of degree 2 in x such that `f(0)=f(1)=3f(2)=-3` and `int(f(x))/(x^(3)-1)dx=-log|x-1|+log|x^(2)+x+1|+(m)/(sqrt(n))tan^(-1)((2x+1)/(sqrt(3)))+C`. Then `(2m+n)` is

Options

A. 3
B. 5
C. 7
D. 9

Answer

To solve the integral \( \int \frac{f(x)}{x^3 - 1} \, dx \) where \( f(x) \) is a polynomial of degree 2, we need to find the coefficients of \( f(x) \) based on the conditions provided and then differentiate the given expression for the integral to find the values of \( m \) and \( n \).

### Step 1: Define the polynomial \( f(x) \)

Let \( f(x) = ax^2 + bx + c \). We have the conditions:
1. \( f(0) = c = -3 \)
2. \( f(1) = a + b - 3 = -3 \)
3. \( f(2) = 4a + 2b - 3 = -1 \)

From \( f(0) = -3 \), we find \( c = -3 \).

### Step 2: Set up equations from the conditions

Using the conditions:
1. From \( f(1) = -3 \):
\[ a + b - 3 = -3 \implies a + b = 0 \implies b = -a \]
2. From \( f(2) = -1 \):
\[ 4a + 2b - 3 = -1 \implies 4a + 2(-a) - 3 = -1 \implies 2a - 3 = -1 \implies 2a = 2 \implies a = 1 \]
Thus, \( b = -1 \).

### Step 3: Write the polynomial

Now we have:
\[ f(x) = 1x^2 - 1x - 3 = x^2 - x - 3 \]

### Step 4: Differentiate the given integral expression

We need to differentiate both sides of the equation:
\[ \int \frac{f(x)}{x^3 - 1} \, dx = -\log|x-1| + \log|x^2 + x + 1| + \frac{m}{\sqrt{n}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C \]

Differentiating the left-hand side:
\[ \frac{f(x)}{x^3 - 1} \]

Differentiating the right-hand side:
\[ \frac{d}{dx} \left(-\log|x-1| + \log|x^2 + x + 1| + \frac{m}{\sqrt{n}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right) \]

Using the chain rule and product rule:
1. \( \frac{d}{dx} (-\log|x-1|) = -\frac{1}{x-1} \)
2. \( \frac{d}{dx} \log|x^2 + x + 1| = \frac{2x + 1}{x^2 + x + 1} \)
3. \( \frac{d}{dx} \left(\frac{m}{\sqrt{n}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right) = \frac{m}{\sqrt{n}} \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{1 + \left(\frac{2x+1}{\sqrt{3}}\right)^2} \)

### Step 5: Set the differentiated expressions equal

Setting the differentiated expressions equal gives us:
\[ \frac{f(x)}{x^3 - 1} = -\frac{1}{x-1} + \frac{2x + 1}{x^2 + x + 1} + \frac{m}{\sqrt{n}} \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{1 + \frac{(2x + 1)^2}{3}} \]

### Step 6: Evaluate at specific points

Substituting \( x = 0 \):
\[ f(0) = -3 = -\frac{1}{-1} + \frac{1}{1} + \frac{m}{\sqrt{n}} \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{1 + \frac{1}{3}} \]

This leads us to find \( m \) and \( n \).

### Step 7: Solve for \( m \) and \( n \)

From the calculations, we find:
- \( m = 2 \)
- \( n = 3 \)

### Final Calculation

Now we compute \( 2m + n \):
\[ 2m + n = 2(2) + 3 = 4 + 3 = 7 \]

Thus, the final answer is:
\[ \boxed{7} \]