I was searching the internet when I came a across a question, and just couldn't solve it. I kept rearranging and substituting but kept going around in loops.
"For $n:= 1,2,3,.....,$ Let $$ I_n = \int_0^1 \frac{x^{n-1}}{2-x}.dx $$ Writing $x^n =x^{n-1}(2-(2-x))$, show that this sequence of numbers satisfies the recurrence relation $$I_{n+1} = 2I_n - \frac{1}{n} $$
Extension
The value taken for $I_1 = \ln{2}$ is $0.6931 $ if the recurrence relation is now used to calculate successive values for $I_n$ we find $I_{12}=-0.0189$ (you are not required to confirm the calculation). Explain carefully both why this cannot be correct and the source of the error, given that all the intermediate arithmetical operations have been performed exactly.
Well, $$\begin{align}I_{n+1} &= \int_0^1\frac{x^n}{2-x}\,dx\\ &= \int_0^1\frac{x^{n-1}\bigl(2-(2-x)\bigr)}{2-x}\,dx\\ &= \int_0^1\frac{2x^{n-1}-(2-x)x^{n-1}}{2-x}\,dx\\ &= \int_0^1\left(\frac{2x^{n-1}}{2-x}-\frac{(2-x)x^{n-1}}{2-x}\right)\,dx.\end{align}$$ Can you take it from there?