In the section on Taylor's theorem (6.4) in Introduction to Real Analysis by Bartle (4th Edition), it says 
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Here $R_n(x)$ is the Lagrange remainder for $ln(1+x)$ expanded around $x_0=0$. I do not understand how to arrive at it. I computed it using the expression $$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}$$ to get $\frac{(-1)^n}{n+1}\frac{1}{(1+c)^{n+1}}x^{n+1}$ where $0 < c < x$. I don't see how to go from my form of $R_n(x)$ to the book's. I tried using $\frac{1}{1+c}$ as a new $c$, but the range becomes a problem if $x \leq 1$. Any hints will help.
The book's version seems incorrect. Take $n=1$ and $x=0.1$. Then $\ln(1+x)\approx 0.09531$ and $P_n(x)=x=0.1$, so $R_1(x)\approx-0.00469.$ If we then use the book's form we have $R_1(x)=-c^2 x^2/2\implies c\approx 0.97$ which is bigger than $x=0.1$. By contrast, your form $R_1(x)=-x^2/(2(1+c)^2)$ yields $c\approx 0.0325$ which is smaller than $0.1$. So I would chalk it up to an error on the book's part. If I were to speculate, the error was missed because the argument was only being used to determine the sign of the remainder; the book's version does give this correctly, so the fact that it's the wrong remainder term was not noticed.