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

y = ln(x+(√(x^2 +1))) (d^n y/dx^n ) = ?

$$ \\ $$$$\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\:=\:? \\ $$$$ \\ $$$$ \\ $$

Answered by qaz last updated on 17/Jan/23

y=ln(x+(√(x^2 +1)))=∫_0 ^x (dt/( (√(t^2 +1))))=Σ_(k=0) ^∞ (((−1/2)),(k) )∫_0 ^x t^(2k) dt =Σ_(k=0) ^∞ (((2k)),(k) )(−(1/4))^k (x^(2k+1) /(2k+1)) ⇒y^((n)) =Σ_(k=0) ^∞ (((2k)),(k) )(−(1/4))^k (((2k+1)(2k)(2k−1)...(2k−n+2))/(2k+1))x^(2k−n+1)

$${y}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{−\mathrm{1}/\mathrm{2}}\\{{k}}\end{pmatrix}\int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{2}{k}} {dt} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix}\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} \frac{{x}^{\mathrm{2}{k}+\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\Rightarrow{y}^{\left({n}\right)} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix}\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} \frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)...\left(\mathrm{2}{k}−{n}+\mathrm{2}\right)}{\mathrm{2}{k}+\mathrm{1}}{x}^{\mathrm{2}{k}−{n}+\mathrm{1}} \\ $$

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