`I= int dx/(x^2+x+1)^2=Atan^-1((2x+1)/(sqrt3))+B(2x+1)/(x^2+x+1)` then Find the value of A and B

Answer

To solve the integral \( I = \int \frac{dx}{(x^2 + x + 1)^2} \) and express it in the form \( I = A \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + B \frac{2x + 1}{x^2 + x + 1} \), we need to find the values of \( A \) and \( B \).

### Step-by-step Solution:

1. **Differentiate Both Sides**:
Start by differentiating both sides of the equation with respect to \( x \):
\[ \frac{dI}{dx} = \frac{1}{(x^2 + x + 1)^2} \]
For the right-hand side:
\[ \frac{d}{dx}\left(A \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + B \frac{2x + 1}{x^2 + x + 1}\right) \]

2. **Differentiate the First Term**:
Using the chain rule for the first term:
\[ \frac{d}{dx}\left(A \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)\right) = A \cdot \frac{1}{1 + \left(\frac{2x + 1}{\sqrt{3}}\right)^2} \cdot \frac{2}{\sqrt{3}} = \frac{2A}{\sqrt{3} \left(1 + \frac{(2x + 1)^2}{3}\right)} \]
Simplifying the denominator:
\[ 1 + \frac{(2x + 1)^2}{3} = \frac{3 + (2x + 1)^2}{3} = \frac{3 + 4x^2 + 4x + 1}{3} = \frac{4x^2 + 4x + 4}{3} = \frac{4(x^2 + x + 1)}{3} \]
Thus,
\[ \frac{d}{dx}\left(A \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)\right) = \frac{2A \cdot 3}{\sqrt{3} \cdot 4(x^2 + x + 1)} = \frac{3A}{2\sqrt{3}(x^2 + x + 1)} \]

3. **Differentiate the Second Term**:
For the second term, apply the quotient rule:
\[ \frac{d}{dx}\left(B \frac{2x + 1}{x^2 + x + 1}\right) = B \cdot \frac{(x^2 + x + 1)(2) - (2x + 1)(2x + 1)}{(x^2 + x + 1)^2} \]
Simplifying the numerator:
\[ 2(x^2 + x + 1) - (2x + 1)^2 = 2x^2 + 2x + 2 - (4x^2 + 4x + 1) = -2x^2 - 2x + 1 \]
Thus,
\[ \frac{d}{dx}\left(B \frac{2x + 1}{x^2 + x + 1}\right) = \frac{B(-2x^2 - 2x + 1)}{(x^2 + x + 1)^2} \]

4. **Combine Derivatives**:
Now equate the derivatives:
\[ \frac{1}{(x^2 + x + 1)^2} = \frac{3A}{2\sqrt{3}(x^2 + x + 1)} + \frac{B(-2x^2 - 2x + 1)}{(x^2 + x + 1)^2} \]

5. **Clear the Denominator**:
Multiply through by \( (x^2 + x + 1)^2 \):
\[ 1 = \frac{3A}{2\sqrt{3}}(x^2 + x + 1) + B(-2x^2 - 2x + 1) \]

6. **Set Up Equations**:
Rearranging gives:
\[ 1 = \left(-2B + \frac{3A}{2\sqrt{3}}\right)x^2 + \left(-2B + \frac{3A}{2\sqrt{3}}\right)x + (B + \frac{3A}{2\sqrt{3}}) \]
For this to hold for all \( x \), coefficients must match:
- Coefficient of \( x^2 \): \( -2B + \frac{3A}{2\sqrt{3}} = 0 \)
- Coefficient of \( x \): \( -2B + \frac{3A}{2\sqrt{3}} = 0 \)
- Constant term: \( B + \frac{3A}{2\sqrt{3}} = 1 \)

7. **Solve the System**:
From \( -2B + \frac{3A}{2\sqrt{3}} = 0 \):
\[ B = \frac{3A}{4\sqrt{3}} \]
Substitute into the constant term equation:
\[ \frac{3A}{4\sqrt{3}} + \frac{3A}{2\sqrt{3}} = 1 \]
Combine terms:
\[ \frac{3A}{4\sqrt{3}} + \frac{6A}{4\sqrt{3}} = 1 \implies \frac{9A}{4\sqrt{3}} = 1 \implies A = \frac{4\sqrt{3}}{9} \]
Substitute \( A \) back to find \( B \):
\[ B = \frac{3 \cdot \frac{4\sqrt{3}}{9}}{4\sqrt{3}} = \frac{4}{9} \]

### Final Values:
Thus, the values of \( A \) and \( B \) are:
\[ A = \frac{4\sqrt{3}}{9}, \quad B = \frac{4}{9} \]