I have this expression: $I_{n} = \int_0^1 \frac{x^{n-1}}{2-x} dx$ for $n=1,2,3,\ldots$ I have been asked to show that by writing $x^n = x^{n-1} (2-(2-x))$ that the recurrence relation $I_{n+1} = 2I_n - \frac{1}{n} $ After hours of rearranging and substituting I keep going round in circles. Has anyone got any ideas of what I need to do?
2026-05-15 22:23:45.1778883825
recurrence relation question.
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$$I_{n+1}-2I_n=\int_0^1\frac{x^{n}-2x^{n-1}}{2-x}dx=-\int_0^1\frac{x^{n-1}(2-x)}{2-x}dx=-\int_0^1 x^{n-1}dx\text{ as }2-x\ne0$$
Use $$\int x^m dx=\begin{cases} \ln x &\mbox{if } m =-1 \\ \frac{x^{m+1}}{m+1}& \mbox{else where }\end{cases}$$