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Question Number 47101 by Meritguide1234 last updated on 04/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

$$\int \frac{dx}{{(1+x^2)}^n}$$$$=\frac{1}{2(n-1)}\times \frac{x}{{(x^2+1)}^{n-1}}+\frac{2n-3}{2n-2}\int \frac{dx}{{(1+x^2)}^{n-1}}$$$$I_n={\int }_0^1\frac{dx}{{(1+x^2)}^n}$$$$=\frac{1}{2(n-1)}\mid \frac{x}{{(1+x^2)}^{n-1}}{\mid }_0^1+\frac{2n-3}{2n-2}{I}_{n-1}$$$$=\frac{1}{2(n-1)}\times \frac{1}{2^{n-1}}+\frac{2n-3}{2n-2}{I}_{n-1}$$$$I_n=\frac{1}{(n-1)}\times \frac{1}{2^n}+\frac{2n-3}{2n-2}{I}_{n-1}$$$$I_1={\int }_0^1\frac{dx}{{(1+x^2)}^1}=\frac{\pi }{4}$$$$I_n=\frac{1}{(n-1)}\times \frac{1}{2^n}+\frac{2n-3}{2n-2}{I}_{n-1}$$$$I_2=\frac{1}{2^2}+\frac{1}{2}\times \frac{\pi }{4}=\frac{1}{2^2}+\frac{\pi }{8}$$$$I_3=\frac{1}{2}\times \frac{1}{2^3}+\frac{3}{4}(\frac{1}{2^2}+\frac{\pi }{8})$$$$=\frac{1}{2^4}+\frac{3}{2^4}+\frac{3\pi }{32}$$$$contd...$$$$$$

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