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Question Number 185086 by SEKRET last updated on 16/Jan/23

$$$$$$$$$$$$$$\mathbf{y}=\mathbf{ln}(\mathbf{x}+\sqrt{{\mathbf{x}}^2+1})$$$$$$$$\frac{{\mathbf{d}}^{\mathbf{n}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{n}}}=?$$$$$$$$$$

Answered by qaz last updated on 17/Jan/23

$$y=ln(x+\sqrt{x^2+1})={\int }_0^x\frac{dt}{\sqrt{t^2+1}}=\underset{k=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}-1/2\\ k\end{array}\right){\int }_0^x{t}^{2k}dt$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}2k\\ k\end{array}\right){(-\frac{1}{4})}^k\frac{x^{2k+1}}{2k+1}$$$$\Rightarrow y^{\left(n\right)}=\underset{k=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}2k\\ k\end{array}\right){(-\frac{1}{4})}^k\frac{(2k+1)\left(2k\right)(2k-1)...(2k-n+2)}{2k+1}{x}^{2k-n+1}$$

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