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$$\ln (x+\sqrt{x^2+1})$$

I don't know how to proceed in a way other than finding the successive derivatives of this function.

First, I was trying to make the substitution $1 + t = x+\sqrt{x^2+1}$ but after writing the Maclaurin Series I noticed that in that way I wasn't taking into account the derivatives in terms of $x$.

Then I tried the following

$$x+\sqrt{x^2+1} = x + 1 + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}x^{2k} + o(x^{2n})$$

If I put

$$t' = x + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}x^{2k} + o(x^{2n})$$

the logarithm takes the form $\ln(1+t')$, but if I tried to expand this in Maclaurin Series I'll get a quite messy expression in powers of $t'$, which doesn't allow me to write the required expression in closed form (or maybe I don't have the knowledge to do that from it).

asd
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    Hint: $$ \frac{d}{dx}\log(x+\sqrt{x^2+1})=\frac{1}{\sqrt{x^2+1}}$$.Then use the binomial expansion of the derivative and finish by integrating term by term. – Mark Viola Nov 19 '16 at 21:21
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    The derivative of $\ln(x+\sqrt{x^2+1})$ is $\frac{1}{\sqrt{x^2+1}}$. Get higher derivatives and get a pattern. – Isko10986 Nov 19 '16 at 21:21
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    A remark: $ln(x + \sqrt{1 + x^2})$ is identical to $arc\sinh(x)$ (the reciprocal function of $\sinh.$) – Jean Marie Nov 19 '16 at 21:33

1 Answers1

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Hint

If we put $f(x)=\ln(x+\sqrt{1+x^2})$

then

$$f'(x)=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}$$

$$=(1+x^2)^{-\frac 12}$$

$$=1-\frac 12 x^2+\frac 38 x^4-....$$

and integrate to get

$$f(x)=x-\frac 16 x^3+\frac{3}{40} x^5-...$$

hamam_Abdallah
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