Question

Let \(I_n = \int_{0}^{1} \frac{dx}{(1+x^2)^n} \)
Prove that for every positive integer n.
\(2nI_{n+1} = 2^{-n}+(2n-1)I_n \)

Answer 1

@ganeshie8

Answer 2

@IrishBoy123

Answer 3

I tried multiplying the integral by 1 and then using integration by parts bu ended with
\[ I_n = 2^{-n} +2n \int_{0}^{1} \frac{x^2}{(1+x^2)^{n+1}} dx\]
I then tried making the numerator 1+x^2-x^2 and ended up with
\[ I_n = I_{n-1} + \int_{0}^{1} \frac{x^2}{(1+x^2)^{n}} dx\]

What should I do?

Answer 4

\[ I_n = 2^{-n} +2n \int_{0}^{1} \frac{x^2}{(1+x^2)^{n+1}} dx\]
Looks good till above step

Answer 5

yeah but there is a x^2 what should I do with it?

Answer 6

Next you may replace x^2 by 1+x^2-1

Answer 7

ohhhhhhh

Answer 8

gonna kill myself

Answer 9

happens..

Answer 10

thank you very much

Answer 11

Np