Let `A = int _(3//4) ^(4//3) (2x ^(2) + x +1 )/( x ^(3) + x ^(2) + x + 1 ) dx` then find the value of `e ^(A).`

Answer

To solve the integral \( A = \int_{3/4}^{4/3} \frac{2x^2 + x + 1}{x^3 + x^2 + x + 1} \, dx \), we will follow these steps:

### Step 1: Simplify the Denominator
The denominator can be factored as follows:
\[ x^3 + x^2 + x + 1 = (x^2 + 1)(x + 1) \]
This is done by factoring out \( x + 1 \) from the polynomial.

**Hint:** Look for common factors in the polynomial to simplify it.

### Step 2: Rewrite the Integral
Now we can rewrite the integral as:
\[ A = \int_{3/4}^{4/3} \frac{2x^2 + x + 1}{(x + 1)(x^2 + 1)} \, dx \]

**Hint:** Always express the integrand in a simpler form before proceeding with integration.

### Step 3: Use Partial Fraction Decomposition
We can express the integrand using partial fractions:
\[ \frac{2x^2 + x + 1}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \]
Here, \( A \), \( B \), and \( C \) are constants to be determined.

**Hint:** Set up the equation by multiplying both sides by the denominator to eliminate the fraction.

### Step 4: Find Coefficients
Multiplying through by \( (x + 1)(x^2 + 1) \) gives:
\[ 2x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 1) \]
Expanding and comparing coefficients will yield a system of equations:
1. \( A + B = 2 \)
2. \( C + A = 1 \)
3. \( B = 1 \)

From these equations, we can solve for \( A \), \( B \), and \( C \):
- From \( B = 1 \), substitute into the first equation: \( A + 1 = 2 \) gives \( A = 1 \).
- Substitute \( A = 1 \) into the second equation: \( C + 1 = 1 \) gives \( C = 0 \).

Thus, we have:
\[ A = 1, \quad B = 1, \quad C = 0 \]

**Hint:** Use substitution and elimination methods to solve for the coefficients.

### Step 5: Rewrite the Integral
Now we can rewrite the integral:
\[ A = \int_{3/4}^{4/3} \left( \frac{1}{x + 1} + \frac{x}{x^2 + 1} \right) \, dx \]

**Hint:** Break the integral into simpler parts that can be integrated separately.

### Step 6: Integrate Each Term
Integrate each term separately:
1. \( \int \frac{1}{x + 1} \, dx = \ln |x + 1| \)
2. For \( \int \frac{x}{x^2 + 1} \, dx \), use the substitution \( u = x^2 + 1 \) which gives \( \frac{1}{2} \ln |x^2 + 1| \).

Thus,
\[ A = \left[ \ln |x + 1| + \frac{1}{2} \ln |x^2 + 1| \right]_{3/4}^{4/3} \]

**Hint:** Remember to apply limits after integration.

### Step 7: Evaluate the Limits
Calculate \( A \) at the upper and lower limits:
\[ A = \left( \ln \left( \frac{4}{3} + 1 \right) + \frac{1}{2} \ln \left( \left( \frac{4}{3} \right)^2 + 1 \right) \right) - \left( \ln \left( \frac{3}{4} + 1 \right) + \frac{1}{2} \ln \left( \left( \frac{3}{4} \right)^2 + 1 \right) \right) \]

**Hint:** Substitute the limits carefully and simplify.

### Step 8: Final Calculation
After evaluating and simplifying, we find:
\[ A = \ln \left( \frac{4/3}{3/4} \right) = \ln \left( \frac{16}{9} \right) \]

### Step 9: Find \( e^A \)
Finally, we compute:
\[ e^A = e^{\ln \left( \frac{16}{9} \right)} = \frac{16}{9} \]

### Conclusion
Thus, the value of \( e^A \) is:
\[ \boxed{\frac{16}{9}} \]