# Particular Integral, with sequence

Subject: Seeking Help for a Computer Science Contest - Integral Estimation

Hello everyone,

I hope this message finds you well. I am currently preparing for an ongoing computer science contest, and I have encountered a challenging problem related to estimating integrals. The specific problem involves calculating the integral:

$$I_n = \int_0^1 \frac{x^n}{6 + x - x^2} \,dx$$

The initial values are given as $$I_0 = \frac{2}{3}\ln\left(\frac{3}{2}\right)$$ and $$I_1 = \frac{1}{5}\ln\left(\frac{3}{2}\right)$$.

The task is to show that for $$n \geq 2$$, $$I_n$$ can be expressed through the following recurrence relation:

$$I_n = \alpha I_{n-1} + \beta I_{n-2} + \gamma_n$$

where $$\alpha + \beta$$ are constants to be determined, and $$\gamma_n$$ is a sequence that needs to be explicitly defined.

I've been struggling to prove this recurrence relation, and any guidance or assistance would be greatly appreciated. Additionally, I am curious to determine the values of $$\alpha,\beta$$, and the explicit form of $$\gamma_n \$$ for this recurrence relation.

There is a standard way to approach this sort of problems. Note that \begin{align} 6 I_n + I_{n + 1} - I_{n + 2} &= \int_{0}^1 \frac{6 x^n + x^{n + 1} - x^{n + 2}}{6 + x - x^2} dx \\ &= \int_0^1 \frac{x^n(6 + x - x^2)}{6 + x - x^2} dx\\ &= \int_0^1 x^n dx = \frac{1}{n + 1}.\end{align} Thus we have that $$I_{n + 2} = 6 I_n + I_{n + 1} - \frac{1}{n + 1},$$ i.e. $$\alpha = 1, \beta = 6$$ and $$\gamma_{n+2} = \frac{1}{n+1}.$$ I hope this helps. :)