Calculate $\int_0^{1/10}\sum_{k=0}^9 \frac{1}{\sqrt{1+(x+\frac{k}{10})^2}}dx$ How can we evaluate the following integral:  

$$\int_0^{1/10}\sum_{k=0}^9 \frac{1}{\sqrt{1+(x+\frac{k}{10})^2}}dx$$  


I know basically how to calculate by using the substitution $x=\tan{\theta}$ :
$$\int_0^1 \frac{dx}{\sqrt{1+x^2}}$$
But I cannot find out a way to apply the result to the question.

Put $x+\frac k{10}=\sinh u$ (i.e. $dx=\cosh u$), we have
$$\begin{align}
&{\large\int}_0^{1/10}\sum_{k=0}^9\frac 1{\sqrt{1+\left(x+\frac k{10}\right)^2}}dx\\
&=\sum_{k=0}^9{\large\int}_0^{1/10}\frac 1{\sqrt{1+\left(x+\frac k{10}\right)^2}}dx\\
&=\sum_{k=0}^9{\large\int}_\alpha^\beta\frac 1{\sqrt{1+\sinh^2u}}\;\cosh u\; du
&&\scriptsize  \text{where }\alpha=\sinh^{-1}\frac k{10}, \beta=\sinh^{-1}\frac{k+1}{10}\\
&=\sum_{k=0}^9{\large\int}_\alpha^\beta 1\; du\\
&=\sum_{k=0}^9\; \sinh^{-1}\left(\frac{k+1}{10}\right)-\sinh^{-1}\left(\frac k{10}\right)\\
&=\sinh^{-1}1-\sinh^{-1}0&&\scriptsize\text{by telescoping}\\\\
&=\color{red}{\ln(1+\sqrt{2})}\qquad\blacksquare
\end{align}$$
