The problem involves evaluating the integral:

$\int_0^1 \frac{x+1}{x^2 + x + 6} \, dx$

Step-by-step Solution:

Step 1: Decompose the integral

Split the numerator $x + 1$ into two terms: $\int_0^1 \frac{x + 1}{x^2 + x + 6} \, dx = \int_0^1 \frac{x}{x^2 + x + 6} \, dx + \int_0^1 \frac{1}{x^2 + x + 6} \, dx$

Step 2: Work on the first integral

For $\int_0^1 \frac{x}{x^2 + x + 6} \, dx$, use substitution: Let $u = x^2 + x + 6$. Then, $du = (2x + 1)dx$.

We rewrite: $x \, dx = \frac{du - dx}{2} \quad \text{and } x^2 + x + 6 = u.$

Substitute back and simplify the first term.

Stepfrom sympy import symbols, integrate

Define variables

x = symbols('x')

Define the function

integral_expr = (x + 1) / (x**2 + x + 6)

Compute the integral

integral_result = integrate(integral_expr, (x, 0, 1)) integral_result-log(6)/2 - sqrt(23)*atan(sqrt(23)/23)/23 + sqrt(23)atan(3sqrt(23)/23)/23 + log(8)/2-log(6)/2 - sqrt(23)*atan(sqrt(23)/23)/23 + sqrt(23)atan(3sqrt(23)/23)/23 + log(8)/2successThe evaluated integral is:

$\int_0^1 \frac{x+1}{x^2 + x + 6} \, dx = -\frac{\ln(6)}{2} - \frac{\sqrt{23}}{23} \arctan\left(\frac{\sqrt{23}}{23}\right) + \frac{\sqrt{23}}{23} \arctan\left(\frac{3\sqrt{23}}{23}\right) + \frac{\ln(8)}{2}$

This is the exact result of the integral. Let me know if you'd like further details or a numerical approximation.

Questions to Explore Further:

1. Would you like a step-by-step explanation of how the terms are derived?
2. Should I simplify this expression further for specific values or conditions?
3. Do you want to verify this result numerically?
4. Would you like an explanation of the substitution process used in this solution?
5. Should I plot the integrand to visualize its behavior?

Tip:

When working with integrals involving polynomials in the denominator, always consider substitution or partial fraction decomposition.