How to evaluate this Integral $\int { {\sqrt{5^2+K^2}}dK \over {\sqrt{10^2+K^2}K}} $ While working on an Exact Differential Equation, I encounter the following Integral. 
$$\int { {\sqrt{5^2+K^2}} \over {K\sqrt{10^2+K^2}}} dK$$
I have tried substitution and all the other elementary methods, but the Integral simply refuses to yield to all my attempts. I reckon that this requires some concept that I've yet not studied. Maybe elliptical Integrals? But that's just speculation. Please Help....
Thank You.
 A: This is not a trivial one $$I=\int\frac{\sqrt{x^2+25}}{x \sqrt{x^2+100}}\,dx$$ Let us try using $$\frac{\sqrt{x^2+25}}{ \sqrt{x^2+100}}=u^2 \implies x=\frac{5 \sqrt{1-4 u^4}}{\sqrt{u^4-1}}\implies dx= \frac{30 u^3}{\sqrt{1-4 u^4} \left(u^4-1\right)^{3/2}}du$$ So, $$I=-\int\frac{6 u^5}{4 u^8-5 u^4+1}\,du$$ Now, since the denominator shows pretty nice roots (it is a quadratic in $u^4$), partial fraction decomposition leads to $$\frac{-6 u^5}{4 u^8-5 u^4+1}=\frac{u}{u^2+1}+\frac{u}{2 u^2-1}-\frac{u}{2 u^2+1}-\frac{1}{2 (u-1)}-\frac{1}{2
   (u+1)}$$ makes the problem much more pleasant.
The result of the integration is just a sum of logarithms $$I=\frac{1}{4} \log \left(1-2 u^2\right)-\frac{1}{2} \log
   \left(1-u^2\right)+\frac{1}{2} \log \left(1+u^2\right)-\frac{1}{4} \log \left(1+2
   u^2\right)$$ Recombining all of that simplifies again and $$I=\tanh ^{-1}\left(u^2\right)-\frac{1}{2} \tanh ^{-1}\left(2 u^2\right)$$
Edit
Applying the same method to
$$I=\int\frac{\sqrt{x^2+a^2}}{x \sqrt{x^2+b^2}}\,dx$$  would lead to $$I=\tanh ^{-1}\left(u^2\right)-\frac{a }{b}\tanh ^{-1}\left(\frac{b }{a}u^2\right)$$
