If `I_n=int_0^1(dx)/((1+x^2)^n); n in N ,` then prove that `2nI_(n+1)=2^(-n)+(2n-1)I_n`

Options

A. `2nI_(n+1)=2^(-n)+(2n-1)I_(n)`
B. `I_(2)=(pi)/8+1/4`
C. `I_(2)=(pi)/8-1/4`
D. `I_(3)=(3pi)/32+1/4`

Answer

A, B, D

Explanation

`I_(n)=int_(0)^(1)(dx)/((1+x^(2))^(n))=int_(0)^(1)(1+x^(2))^(-n)dx`
`=x/((1+x^(2))^(n))|_(0)^(1)-int_(0)^(1)(-n)(1+x^(2))^(-n-1)2x.xdx`
`=1/(2^(n))+2nint_(0)^(1)(x^(2)dx)/((1+x^(2))^(n+1))`
`=1/(2^(n))+2n int_(0)^(1)(1+x^(2)-1)/((1+x^(2))^(n+1))dx`
`=1/(2^(n))+2nI_(n)-2nI_(n+1)`
or `2nI_(n+1)=2^(-n)+(2n-1)I_(n)`
or `2I_(2)=1/2+I_(1)=1/2+tan^(-1)x|_(0)^(1)`
or `I_(2)=1/4+(pi)/8`
Also `4I_(3)=2^(-2)+3I_(2)=1/4+3(1/4+(pi)/8)`.so `I_(3)=(3pi)/32`