Reoccurance Integration Question (1 Viewer)

[sasquatch]

Could somebody please help me with this question? Ive tried rewriting it as (1+x)^n-1 * (1+x) for the denominator and alot of other variants... Cant get the answer !

If I_n = ₀∫^t 1/(1+x²)ⁿ dx) where n >= 1, show that:

2nI_n+1 = (2n-1)I_n + t/(1+t²)ⁿ.

Thanks in advance.

 

[Yip]

I_n = ₀∫^t 1/(1+x²)ⁿ dx)
Integrating by parts,
I_n =[x/(1+x²)ⁿ]₀t -₀∫^t x.[-2nx(1+x²)^n-1/ (1+x²)²ⁿ] dx
= t/(1+t²)ⁿ+2n₀∫^t [x²/ (1+x²)ⁿ⁺¹] dx
=t/(1+t²)ⁿ+2n₀∫^t [(x²+1-1)/ (1+x²)ⁿ⁺¹] dx
=t/(1+t²)ⁿ+2n[I_n -I_n+1]

故に

2nI_n+1 = (2n-1)I_n + t/(1+t²)ⁿ

ὅπερ ἔδει δεῖξαι

 

[sasquatch]

Woah... i really dont understand that..hehehe. Thanks though ill look at it and try to figure it out..