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.
 A: There is a better way.  
We shall prove that:  

$$\int_0^s f(x+ks)dx=\int_{ks}^{(k+1)s}f(x)dx \tag1$$  

And hence:  

$$\int_0^s[f(x)+f(x+s)+...+f(x+(n-1)s)]dx=\int_0^{ns}f(x)dx \tag2$$

Proof: 
Using substitution $t\mapsto x+ks$,
$$\int_0^s f(x+ks)dx=\int_{ks}^{(k+1)s}f(t)dt$$
\begin{align}
&   \int_0^s[f(x)+f(x+s)+\cdots+f(x+(n-1)s)]dx \\
& = \int_0^sf(x)dx+\int_0^sf(x+s)dx+\cdots+\int_0^sf(x+(n-1)s)dx \\
& = \int_0^sf(x)dx+\int_s^{2s}f(x)dx+\cdots+\int_{(n-1)s}^{ns}f(x)dx \\
& = \int_0^{ns}f(x)dx
\end{align}  

Using the aforementioned results, your integral just becomes:
$$\int_0^1\frac{dx}{\sqrt{1+x^2}} \tag3$$
, which is exactly equal to your given integral!
A: Consider $$I_k=\int \frac{dx}{\sqrt{1+(x+\frac{k}{10})^2}}$$ and let $x+\frac{k}{10}=\sinh(y)$, $dx=\cosh(y)\, dy$ which make $$I_k=\int dy=y=\sinh^{-1} \left(x+\frac k{10}\right)$$ So,$$J_k=\int_0^{\frac 1{10}} \frac{dx}{\sqrt{1+(x+\frac{k}{10})^2}}=\sinh ^{-1}\left(\frac{k+1}{10}\right)-\sinh ^{-1}\left(\frac{k}{10}\right)$$ which leads to nicely telescoping terms.
I am sure that you can take it from here.
A: Put $x+\frac k{10}=\sinh u$ (i.e. $dx=\cosh u$), we have
$$\begin{align}
&{\large\int}_0^{1/10}\sum_{k=0}^9\frac 1{\sqrt{1+\left(x+\frac k{10}\right)^2}}dx\\
&=\sum_{k=0}^9{\large\int}_0^{1/10}\frac 1{\sqrt{1+\left(x+\frac k{10}\right)^2}}dx\\
&=\sum_{k=0}^9{\large\int}_\alpha^\beta\frac 1{\sqrt{1+\sinh^2u}}\;\cosh u\; du
&&\scriptsize  \text{where }\alpha=\sinh^{-1}\frac k{10}, \beta=\sinh^{-1}\frac{k+1}{10}\\
&=\sum_{k=0}^9{\large\int}_\alpha^\beta 1\; du\\
&=\sum_{k=0}^9\; \sinh^{-1}\left(\frac{k+1}{10}\right)-\sinh^{-1}\left(\frac k{10}\right)\\
&=\sinh^{-1}1-\sinh^{-1}0&&\scriptsize\text{by telescoping}\\\\
&=\color{red}{\ln(1+\sqrt{2})}\qquad\blacksquare
\end{align}$$
