Let I_n = ₀^1 1 (x^2+1)^n , dx for n 1 . If I₃ = a + b 32 , where a and b are integers, then the value of a+b is

Step-by-step solution

Given, I_n = ₀^1 1 (x^2+1)^n , dx Applying integration by parts by taking u = 1 (x^2+1)^n and dv = dx , we get: I_n = [ x (x^2+1)^n ]₀^1 - ₀^1 x ( -2nx (x^2+1)^ n+1 ) , dx I_n = 1 2^n + 2n ₀^1 x^2 (x^2+1)^ n+1 , dx Adding and subtracting 1 in the numerator of the integral: I_n = 1 2^n + 2n ₀^1 (x^2+1) - 1 (x^2+1)^ n+1 , dx I_n = 1 2^n + 2n ₀^1 1 (x^2+1)^n , dx - 2n ₀^1 1 (x^2+1)^ n+1 , dx I_n = 1 2^n + 2n I_n - 2n I_ n+1 Rearranging the terms, we obtain the reduction formula: 2n I_ n+1 = (2n-1) I_n + 1 2^n For n=1