If I_n = int_{0}^{1} frac{dx}{(1 + x^2)^n}; n | vedclass

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(D) Let $I_n = \int_{0}^{1} (1 + x^2)^{-n} dx$. Using integration by parts,let $u = (1 + x^2)^{-n}$ and $dv = dx$. Then $du = -n(1 + x^2)^{-n-1} (2x)

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(D) Let $I_n = \int_{0}^{1} (1 + x^2)^{-n} dx$. Using integration by parts,let $u = (1 + x^2)^{-n}$ and $dv = dx$. Then $du = -n(1 + x^2)^{-n-1} (2x) dx$ and $v = x$.$I_n = [x(1 + x^2)^{-n}]_{0}^{1} - \int_{0}^{1} x \cdot (-2nx)(1 + x^2)^{-n-1} dx$$I_n = \frac{1}{2^n} + 2n \int_{0}^{1} \frac{x^2}{(1 + x^2)^{n+1}} dx$$I_n = \frac{1}{2^n} + 2n \int_{0}^{1} \frac{(1 + x^2) - 1}{(1 + x^2)^{n+1}} dx$$I_n = \frac{1}{2^n} + 2n [I_n - I_{n+1}]$$2n I_{n+1} = \frac{1}{2^n} + (2n - 1) I_n$. This confirms statement $(A)$.For $n=1$,$I_1 = \int_{0}^{1} \frac{dx}{1+x^2} = [	an^{-1} x]_{0}^{1} = \frac{\pi}{4}$.Using the reduction formula for $n=1$: $2(1) I_2 = \frac{1}{2^1} + (2(1) - 1) I_1 \implies 2 I_2 = \frac{1}{2} + \frac{\pi}{4} \implies I_2 = \frac{1}{4} + \frac{\pi}{8}$. This confirms statement $(B)$.Thus,both $(A)$ and $(B)$ are correct.

If $I_n = \int_{0}^{1} \frac{dx}{(1 + x^2)^n}$; $n \in N$,then which of the following statements hold good?

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