If I_n = int frac{1}{(x^2+1)^n} dx, then 2n I | vedclass

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(B) We have $I_n = \int \frac{1}{(x^2+1)^n} dx$. Consider $I_n = \int 1 \cdot (x^2+1)^{-n} dx$. Using integration by parts, let $u = (x^2+1)^{-n}$ and

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$\frac{x}{(x^2+1)^n} + c$

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(B) We have $I_n = \int \frac{1}{(x^2+1)^n} dx$. Consider $I_n = \int 1 \cdot (x^2+1)^{-n} dx$. Using integration by parts, let $u = (x^2+1)^{-n}$ and $dv = dx$. Then $du = -n(x^2+1)^{-n-1} \cdot 2x dx = -2nx(x^2+1)^{-n-1} dx$ and $v = x$. $I_n = x(x^2+1)^{-n} - \int x \cdot (-2nx)(x^2+1)^{-n-1} dx$. $I_n = \frac{x}{(x^2+1)^n} + 2n \int \frac{x^2}{(x^2+1)^{n+1}} dx$. $I_n = \frac{x}{(x^2+1)^n} + 2n \int \frac{(x^2+1)-1}{(x^2+1)^{n+1}} dx$. $I_n = \frac{x}{(x^2+1)^n} + 2n \left[ \int \frac{1}{(x^2+1)^n} dx - \int \frac{1}{(x^2+1)^{n+1}} dx \right]$. $I_n = \frac{x}{(x^2+1)^n} + 2n I_n - 2n I_{n+1}$. Rearranging the terms, we get $2n I_{n+1} = \frac{x}{(x^2+1)^n} + (2n-1) I_n$. Therefore, $2n I_{n+1} - (2n-1) I_n = \frac{x}{(x^2+1)^n} + c$.

If $I_n = \int \frac{1}{(x^2+1)^n} dx$, then $2n I_{n+1} - (2n-1) I_n = $

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