Arc Length of $\ln\left(x\right)$

$\begin{aligned}\text{Arc length}&=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\ f(n)&=\int_1^n\sqrt{1+\left(\frac{d}{dx}\ln x\right)^2}dx\\ &=\int_1^n\sqrt{1+\left(\frac{1}{x}\right)^2}dx\\ &=\int_1^n \frac{1}{x}\sqrt{x^2+1}dx\\ \text{let }x=\tan\theta,&\quad\therefore dx=\sec^2\theta d\theta\\ f(n)&=\int_{\tan^{-1}(1)}^{\tan^{-1}(n)} \frac{\sec^2\theta}{\tan\theta}\sqrt{\tan^2\theta+1}d\theta\\ &=\int_\frac{\pi}{4}^{\tan^{-1}(n)} \frac{\sec^2\theta}{\sin\theta\sec\theta}\sqrt{\sec^2\theta}d\theta\\ &=\int_\frac{\pi}{4}^{\tan^{-1}(n)} \sec^2\theta\csc\theta d\theta\\ &=\left[\tan\theta\csc\theta\right]_\frac{\pi}{4}^{\tan^{-1}(n)}-\int_\frac{\pi}{4}^{\tan^{-1}(n)} \tan\theta(-\csc\theta\cot\theta) d\theta\\ &=\left[\sec\theta\right]_\frac{\pi}{4}^{\tan^{-1}(n)}+\int_\frac{\pi}{4}^{\tan^{-1}(n)} \csc\theta d\theta\\ &=\sec\tan^{-1}(n)-\sec\frac{\pi}{4}-\left[\ln(\csc\theta+\cot\theta)\right]_\frac{\pi}{4}^{\tan^{-1}(n)}\\ &=\sqrt{1+n^2}-\sqrt2-\ln\left[\csc\tan^{-1}(n)+\cot\tan^{-1}(n)\right]+\ln\left(\csc\frac{\pi}{4}+\cot\frac{\pi}{4}\right)\\ &=\sqrt{1+n^2}-\sqrt2-\ln\left(\frac{\sqrt{1+n^2}}{n}+\frac{1}{n}\right)+\ln\left(\sqrt2+1\right)\\ \text{Now, }\lim_{n\to\infty}\left(n-f(n)\right)&=\lim_{n\to\infty}\left[n-\sqrt{1+n^2}+\sqrt2+\ln\left(\frac{\sqrt{1+n^2}}{n}-\frac{1}{n}\right)-\ln\left(\sqrt2+1\right)\right]\\ &=\lim_{n\to\infty}\left[\frac{n^2-(1+n^2)}{n+\sqrt{1+n^2}}+\sqrt2+\ln\left(\sqrt{\frac{1}{n^2}+1}-\frac{1}{n}\right)+\ln\left(\frac{1}{\sqrt2+1}\right)\right]\\ &=\lim_{n\to\infty}\left[\frac{1}{n+\sqrt{1+n^2}}+\ln\left(\sqrt{\frac{1}{n^2}+1}-\frac{1}{n}\right)+\ln\left(\frac{\sqrt2-1}{2-1}\right)+\sqrt2\right]\\ &=0+\ln\left(\sqrt{0+1}-0\right)+\ln\left(\frac{\sqrt2-1}{2-1}\right)+\sqrt2\\ &=\ln(1)+\ln\left(\sqrt2-1\right)+\sqrt2\\ &=\ln\left(\sqrt2-1\right)+\sqrt2\end{aligned}$ So the answer is $2+1+2+0=\boxed{5}$