Wolfram Alpha can't solve this integral analytically Wolfram Alpha isn't able to calculate this integral (I don't have mathematica, but I have Wolfram Pro).
$$\int_{0}^{a} \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx \ \ \ , \ b>a$$
This is for a physics problem. I'd appreciate either a solution or the knowledge that the integral is non-soluble (which would indicate that I need to find some symmetry that I haven't seen yet). Thanks!
 A: I was able to specify the closed-form with Maple.
$$\int_{0}^{a} \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx$$
equals to
$$\frac {\sqrt 2}2 \left(\ln  \left(  \left( \sqrt {2}+2\,{\operatorname{csgn}} \left( a-b \right)  \right)  \left( a-b \right)  \right) - \ln  \left(2\,\sqrt {{a}^{2}+{b}^{2}}-\sqrt {2}\left(a+b\right) \right) \right), 
$$
where $\operatorname{csgn}$ is the complex signum function.
If we assume that $b>a$, then we could simplify it into the form
$$\frac {\sqrt 2}2 \left( \ln  \left( b-a \right)+\ln  \left( \sqrt {2}-1 \right) -\ln  \left( \sqrt {2} \sqrt {{a}^{2}+{b}^{2}}-a-b \right) \right).$$
A: Wolfram Alpha evaluates the indefinite integral as
$$\int \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx =\frac{\log\left(\sqrt2 \sqrt{2x^2 - 2(a+b)x + a^2 + b^2} + 2x - a - b \right)}{\sqrt2}+\mbox{constant}.$$
You can get to Maple's closed-form solution from this if you plug in the
limits of the integration, but personally I think the arccosh solution is neater.
A: WolframAlpha give this : http://www.wolframalpha.com/input/?i=integrate+dx%2Fsqrt%28%28x-a%29%5E2%2B%28x-b%29%5E2%29+
$$\int\frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx =\frac{1}{\sqrt 2} \ln\big(\sqrt{2a^2-4ax+2b^2-4bx+4x^2}-a-b+2x\big)+c$$
Then, it is easy to compute the defined integrtal :
$$\int_0^a\frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx =\frac{1}{\sqrt 2} \ln \frac{\sqrt{2a^2+2b^2-4ab}+a-b}{\sqrt{2a^2+2b^2}-a-b}$$
