After reading the reduction formula in the post, I am curious about the closed form of the integral
$$
I_n=\int_0^1 \frac{x^n}{6+x-x^2} d x
$$
I first resolve the integrand into two partial fractions as
$$
I_n=\frac{1}{5} \left[\underbrace{\int_0^1 \frac{x^n}{3-x} d x}_{J_n} +\underbrace{\int_0^1 \frac{x^n}{2+x} d x}_{K_n} \right]
$$
Then I tackle the two integrals in general,
$$
\begin{aligned}
\int_0^1 \frac{x^n}{a-x} d x & =\int_{a-1}^a \frac{(a-x)^n}{x} d x \\
\\
& =\int_{a-1}^a\left(\frac{a^n}{x}-\sum_{k=1}^n\left(\begin{array}{c}
n \\
k
\end{array}\right) a^{n-k} x^{k-1}\right) dx \\
& =a^n \ln \left(\frac{a}{a-1}\right)-\sum_{k=1}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{a^{n}-a^{n-k}(a-1)^k}{k}
\end{aligned}
$$
Hence
$$J_n =3^n \ln \left(\frac{3}{2}\right)-\sum_{k=1}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{3^{n}-3^{n-k}2^k}{k} $$
and
$$K_n= -\left[(-2)^n \ln \left(\frac{2}{3}\right)-\sum_{k=1}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(-2)^{n}-(-1)^n2^{n-k}3^k}{k} \right]$$
We can now conclude that the closed form of the integral is
$$
\begin{aligned}I_n= &\ \frac{1}{5} \left[ 3^n\left[\ln \left(\frac{3}{2}\right)-\sum_{k=1}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{1-\left(\frac{2}{3}\right)^k}{k}\right]+ (-2)^{n}\left[\ln \left(\frac{3}{2}\right)-\sum_{k=1}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{1-\left(\frac{3}{2}\right)^k}{k}\right]\right]\\=& \frac{3^n+(-2)^n}{5} \ln \left(\frac{3}{2}\right) -\frac{1}{5}\left[\sum_{k=1}^n\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{1}{k}\left(3^n\left(1-\left(\frac{2}{3}\right)^k\right)+(-2)^n\left(1-\left(\frac{3}{2}\right)^k\right)\right]\right.
\end{aligned}
$$
My question:
Is there any alternative method? Your comments and alternative methods are highly appreciated.
Best Answer
Here is a method to arrive at a simpler closed-form result. Note that
\begin{align} J_n(a)=&\int_0^1 \frac{x^n}{x+a}dx = \frac1n - a J_{n-1}(a)\\ = &\ (-a)^n\ln\frac{a+1}a+\sum_{k=0}^{n-1}\frac{(-a)^k}{n-k} \end{align} Then \begin{align} &\int_0^1 \frac{x^n}{6+x-x^2} d x\\ =&\ \frac{1}{5} \int_0^1 \frac{x^n}{2+x}-\frac{x^n}{x-3}\ d x = \frac15[J_n(2)-J_n(-3)] \\ =&\ \frac{3^n+(-2)^n}5\ln\frac32 +\frac15 \sum_{k=0}^{n-1}\frac{(-2)^k-3^k}{n-k} \end{align}