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.

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!
