Let I_{n}(x)=int_{0}^{x} frac{1}{(t^{2}+5)^{n | vedclass

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(A) Given $I_{n}(x)=\int_{0}^{x} \frac{1}{(t^{2}+5)^{n}} dt$. By applying integration by parts,let $u = (t^{2}+5)^{-n}$ and $dv = dt$. Then $du = -n(t

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$50 I_{6}-9 I_{5}= x I_{5}^{\prime}$

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(A) Given $I_{n}(x)=\int_{0}^{x} \frac{1}{(t^{2}+5)^{n}} dt$. By applying integration by parts,let $u = (t^{2}+5)^{-n}$ and $dv = dt$. Then $du = -n(t^{2}+5)^{-n-1}(2t) dt$ and $v = t$. $I_{n}(x) = \left[ \frac{t}{(t^{2}+5)^{n}} \right]_{0}^{x} - \int_{0}^{x} t \cdot (-n)(t^{2}+5)^{-n-1}(2t) dt$. $I_{n}(x) = \frac{x}{(x^{2}+5)^{n}} + 2n \int_{0}^{x} \frac{t^{2}}{(t^{2}+5)^{n+1}} dt$. $I_{n}(x) = \frac{x}{(x^{2}+5)^{n}} + 2n \int_{0}^{x} \frac{(t^{2}+5)-5}{(t^{2}+5)^{n+1}} dt$. $I_{n}(x) = \frac{x}{(x^{2}+5)^{n}} + 2n I_{n}(x) - 10n I_{n+1}(x)$. Rearranging the terms,we get $10n I_{n+1}(x) = \frac{x}{(x^{2}+5)^{n}} + (2n-1) I_{n}(x)$. For $n=5$,$50 I_{6}(x) = \frac{x}{(x^{2}+5)^{5}} + 9 I_{5}(x)$. Note that $I_{5}^{\prime}(x) = \frac{d}{dx} \int_{0}^{x} \frac{1}{(t^{2}+5)^{5}} dt = \frac{1}{(x^{2}+5)^{5}}$. Thus,$50 I_{6}(x) = x I_{5}^{\prime}(x) + 9 I_{5}(x)$,which implies $50 I_{6} - 9 I_{5} = x I_{5}^{\prime}$.

Let $I_{n}(x)=\int_{0}^{x} \frac{1}{(t^{2}+5)^{n}} dt, n=1, 2, 3, \ldots$. Then

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