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.

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.
