So here is my solving process: $$ \begin{split} f(x)=x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}&\Rightarrow f'(x)=\ln(x+\sqrt{1+x^2})\\ &\Rightarrow f''(x)=\frac{1}{\sqrt{1+x^2}} \end{split} $$ and from here I apply the generalized binomial formula $(x+y)^r=\sum_{k=0}^\infty\tbinom{r}{k} x^{r-k}y^k$ with $r=-\frac{1}{2}$: $$ f''(x)=\frac{1}{\sqrt{1+x^2}}=1+\sum_{n=1}^\infty\frac{(-1)^n(2n-1)!!}{(2n)!!}x^{2n} $$and integrate $f''(x)$ back to get the expansion series of the original function $$ f'(x)=x+\sum_{n=1}^\infty\frac{(-1)^n(2n-1)!!}{(2n+1)(2n)!!}x^{2n+1} $$ $$ f(x)=\frac{1}{2}x^2+\sum_{n=1}^\infty\frac{(-1)^n(2n-1)!!}{(2n+1)(2n+2)!!}x^{2n+2} $$ But the correct answer should be $f(x)=-1+\frac{1}{2}x^2+\sum_{n=1}^\infty\frac{(-1)^n(2n-1)!!}{(2n+1)(2n+2)!!}x^{2n+2}$. I don't know where the constant term -1 comes from(but it makes sense since $f(0)=-1$), is there any problem in my solution? Or should I always check for constants possibly missing in my series expansion?
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\lninstead of writing justln. Note the difference between$\ln x$and$ln x$, which produce $\ln x$ and $ln x$ respectively. – Theo Bendit Mar 27 '21 at 07:33