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$$I_n = \int\frac{x^{2n}}{(1+x^2)^{1/2}}$$

Having difficulty with, $$x^{2n}$$ oppose to the more common, $$x^{n}$$ which can be split into: $$x^{n-1}*x$$ Used integration by parts to achieve the following by letting: $$u = x^{2n-1}$$ $$du = \frac{x^{2n-2}dx}{2n-2}$$ $$dv = \frac{xdx}{(1+x^2)^{1/2}}$$ $$v = {(x^2+1)^{1/2}}$$ Then started algebra to achieve the following: $$I_n = {x^{2n-1}}{(x^2+1)}^{1/2} - \int\frac{(x^2+1)^{1/2}(x)^{2n-2}dx}{(2n-2)}$$ I know that, $$I_n = \int\frac{x^{2n}}{(1+x^2)^{1/2}}$$ Just wondering if, $$\int\frac{(x^2+1)^{1/2}(x)^{2n-2}dx}{(2n-2)} = I_{n-2}$$ Any help is appreciated, thanks.

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Not quite: $$\int(x^2+1)^{1/2}x^{2n-2}\,dx=\int\frac{(x^2+1) \,x^{2n-2}}{(x^2+1)^{1/2}}\,dx=\int\frac{x^{2n}+x^{2n-2}}{(x^2+1)^{1/2}}\,dx=I_{n}+I_{n-1}$$ Then, taking into account the factor of $\frac1{2n-2}$, you'd get something like $$\frac{2n-1}{2n-2} I_n=x^{2n-1}(x^2+1)^{1/2}-\frac{I_{n-1}}{2n-2}$$

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