Question

If$I_n={\int }_0^1\frac{dx}{{(1+x^2)}^n};nϵN,$ then which of the following statements hold good ?

• A. $2n{l}_{n+1}=2^{-n}+(2n-1)l_n$
• B. $l_2=\frac{\pi }{8}+\frac{1}{4}$
• C. $l_2=\frac{\pi }{8}-\frac{1}{4}$
• D. $l_3=\frac{\pi }{16}+\frac{5}{48}$

Answer 1

Answer: (A), (B)

$2n{l}_{n+1}=2^{-n}+(2n-1)l_n$
$l_2=\frac{\pi }{8}+\frac{1}{4}$

$l_n={\int }_0^1{(1+x^2)}^{-n}dx$

$l_n={[{(1+x^2)}^{-n}\cdot x]}_0^1+n{\int }_0^1\frac{2x^2}{{(1+x^2)}^{n+1}}dx$

$=2^{-n}+2n{\int }_0^1\frac{(x^2+1)-1}{{(x^2+1)}^{n+1}}dx$

$l_n=2^{-n}+2n[l_n-l_{n+1}]$

$2n{l}_{n+1}=2^{-n}+(2n-1)l_n$

now substitute $n=1$

$2l_2=2^{-1}+l_1=\frac{1}{2}+{\int }_0^1\frac{1}{1+x^2}dx=\frac{1}{2}+{\left({\tan }^{-1}x\right)}_0^1$

$l_2=\frac{1}{4}+\frac{\pi }{8}$