Prove that $F_n$ satisfies the recurrence relation $F_n = aF_{n-1} - \frac{1}{n}$ Suppose that $F_n = \int_0^1 \frac{x^n}{a-x}dx$, where $a>2$ and $n=0, 1, 2, 3,...$
Prove that $F_n$ satisfies the recurrence relation $F_n = aF_{n-1} - \frac{1}{n}$
My thought is that this can be proved by using an induction proof. We can assume that $F_n$ is true, then prove that $F_{n+1}$ holds, and conclude the proof.
Inductive Hypothesis: Assume that $F_n$ is true ($F_n = aF_{n-1} - \frac{1}{n}$)
Base Case: We need to solve for $F_0$.
$$F_0 = \int_0^1 \frac{x^n}{a-x} dx
 = -\ln(a-x)|^1_0 = -\ln(a-1) + \ln(a) + C$$
Then show that $F_1$ holds the recurrence relation.
$$F_1 = aF_0 - \frac{1}{n}$$
$$\int_0^1\frac{x}{a-x}dx = a[-\ln(a-1) + \ln(a) + C_1] - \frac{1}{n}$$
$$-a\ln(1-a) + a\ln(-a) - 1 + C_2 = -a \ln(a-1) + a \ln(a)+aC_1 - 1$$
$$-a\ln(1-a)+a\ln(-a)-1 = -a\ln(a-1) + a\ln(a)-1$$
Here's where I run into some issue. They are not quite equal, so did I integrate wrong?
Inductive Step: Since we assume that $F_n$ is true, then we need to prove that $F_{n+1}$ is true.
$$F_{n+1} = aF_{n}-\frac{1}{n}$$
$$\int_0^1\frac{x^{n+1}}{a-x}dx = a \int_0^1 \frac{x^n}{a-x} dx - \frac{1}{n}$$
At this point, I am not sure that how to do this integral, and check for equality.
I'd like to know if I'm going in the right direction with this proof, and if I am, then how to do the integration in the inductive step.
 A: Note that \begin{align*}aF_{n-1} - \frac{1}{n} &= a\int_{0}^{1}\frac{x^{n-1}}{a-x}dx - \frac{1}{n}\\
&=a\left(\int_{0}^{1}\frac{x^{n-1}}{a-x}dx-\frac{1}{an}\right)\\
&= a\left(\int_{0}^{1}\frac{x^{n-1}}{a-x}dx \, - \int_{0}^{1}\frac{x^{n-1}}{a}dx\right)\\
&= a\int_{0}^{1}\left(\frac{x^{n-1}}{a-x} - \frac{x^{n-1}}{a}\right)dx\\
&= a\int_{0}^{1}x^{n-1}\frac{x}{a(a-x)}dx\\
&=F_n\end{align*}
EDIT-2
Remember that when computing definite integrals, there will not be a constant such as $C_{1,2}$ and that when computing, say, $$\int_{0}^{1}\frac{x}{a-x}dx$$ that you dont forget to change the limits of integration according to any substitutions you make:
\begin{align*}
F_1=\int_{0}^{1}\frac{x}{a-x}dx &= \int_{a-1}^{a}\frac{a-x}{x}dx\\
\end{align*}
This will avoid those negative signs you got. 
A: Since $a>2$ we can expand the integrand in a geometric series.
\begin{gather*}
F_{n} = a^{n-1}\int_{0}^{1}\dfrac{\left(\frac{x}{a}\right)^{n}}{1-\frac{x}{a}}\, dx = a^{n-1}\int_{0}^{1}\left(\sum_{k=n}^{\infty}\left(\dfrac{x}{a}\right)^{k}\right)\, dx = a^{n-1}\sum_{k=n}^{\infty}\dfrac{1}{(k+1)a^{k}} =\\[2ex]
a\left(a^{n-2}\sum_{k=n-1}^{\infty}\dfrac{1}{(k+1)a^{k}}-\dfrac{1}{an}\right) =aF_{n-1} - \dfrac{1}{n}.
\end{gather*}
A: Integration by parts ($(x^{n+1})'=(n+1)x^{n}$)
