I am trying to find the integral of
$$f(x)=x^2\ln\left(2\sqrt{\frac{a^2-x^2}{a^2+4x^2}}+\sqrt{\frac{5a^2}{a^2+4x^2}}\right)$$
And I am having no luck with it. Does anyone have any ideas? Is it even possible?
I also need the integral of
$$g(x)=\ln\left(2\sqrt{\frac{a^2-x^2}{a^2+4x^2}}+\sqrt{\frac{5a^2}{a^2+4x^2}}\right)$$
which I also am having trouble with. I'm not sure if these two integrals are solved in a similar way or have vastly different methods of solution.
Give common factor $\frac1{\sqrt{a^2+4x^2}}$ inside the logarithm, and then use the fact that $\ln ab=\ln a+\ln b$ , and that $\ln\frac1{\sqrt t}=-\frac12\ln t$ , in order to arrive at $$I=-\frac12\int[x^2]\ln(a^2+4x^2)dx+\int[x^2]\ln\left(2\sqrt{a^2-x^2}+|a|\sqrt5\right)dx$$ after which you will proceed to integrate by parts (at least in the case where the term $x^2$ is missing from inside the integral sign), then exploiting the derivates of inverse trigonometric functions such as arcsine and arctangent, so as to arrive at results similar to $$\int\ln(1+x^2)dx=2\arctan x+x\Big[\ln(1+x^2)-2\Big]$$ $$\int\ln\left(1+\sqrt{1-x^2}\right)dx=\arcsin x+x\left[\ln\left(1+\sqrt{1-x^2}\right)-1\right]$$ Use Wolfram Alpha to verify the result, by copy-pasting the same code you used inside your post, and then waiting patiently for a few seconds for the result to finally appear.