How to Solve for the Integral of x^(n-1) from 0 to 1?

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Having a bit of trouble, any help will be appreciated.
Thanks.
 
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gomes. said:
Having a bit of trouble, any help will be appreciated.
Thanks.

Looks like you'll probably have to use induction on n.

1. Show that the recurrence relation is true for n = 1. (base step)
2. Assume that, for some n in the natural numbers, your recurrence relation holds. Try to show that n + 1 holds as well. (inductive step).

It's probably going to involve a bit of messy algebra.
 
I_{n+1} = \int_0^1 \frac{x^n}{2-x} dx = \int_0^1 \frac{x^{n-1}(2 - (2-x))}{2-x} dx = 2\int_0^1 \frac{x^{n-1}}{2-x} dx - \int_0^1 x^{n-1} dx

= ...
 
Deadstar said:
\int_0^1 x^{n-1} dx

= ...

Thanks, I've got it now (the 2I(n) ) bit, but how do i get the above part to 1/n?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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