Integrating $\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$ This is a problem that my calculus professor gave to his students many years ago. 

$$\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$$ 

Wolfram doesn't find any solution in terms of standard mathematical functions. I'm sure that this integral has a solution otherwise my professor wouldn't have assigned it.
Could someone help me?
 A: I'm not ready to give the full answer, but I guess, I can help a bit.
Try using formula
$$
\arccos(x) + \arccos(y) =
\arccos\left(xy - \sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right)
$$
(for ref. see Wikipedia).
To do this, write
$$
7 x^2 - \sqrt{49 x^4 -50 x^2 +1} =
x \times 7x - \sqrt{\left(1-x^2\right)\left(1-(7x)^2\right)}
$$
A: Sufficient care has to be taken before the assignment of equality sign.
Using https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values,
$0\le\arccos(x)\le\pi$
$\implies0\le\arccos(x)+\arccos(y)\le2\pi$
$\arccos(x)+\arccos(y)$ will be $$=\arccos(xy-\sqrt{(1-x^2)(1-y^2)})$$ iff $\arccos(x)+\arccos(y)\le\pi$
$\iff\dfrac\pi2-\arcsin(x)+\cdots\le\pi$
$\iff\arcsin(x)\ge-\arcsin(y)=\arcsin(-y)$
As $\arcsin$ is an increasing function, we need $$x\ge-y\iff x+y\ge0$$
Otherwise $$\arccos(x)+\arccos(y)=2\pi-\arccos(xy-\sqrt{(1-x^2)(1-y^2)})$$
Can you identify $y$ here?
See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
A: Cross-posted on AoPS here. Answered by ysharifi. I will just put it here in length.

That is much neater than your other integral. Assuming you want to stay in real numbers, first we need to have $x > 0$ because of the presence of $\ln x.$ Secondly we also need to have 
  $$49x^4-50x^2+1=(1-x^2)(1-49x^2) \ge 0,$$
  and so we have either $0 < x \le \frac{1}{7}$ or $x > 1.$ 
  Thirdly, we also need to have $-1 \le 7x^2-\sqrt{49x^4-50x^2+1} \le 1$ and so $7x^2-1 \le \sqrt{49x^4-50x^2+1},$ which does not hold for $x > 1.$ So the domain of the function you want to integrate is $0 < x \le \frac{1}{7}.$ Now see that 
  $$\cos^{-1}(7x^2-\sqrt { 49x^4-50x^2+1})=\cos^{-1}(7x^2-\sqrt{1-x^2}\sqrt{1-49x^2})$$
$$=\cos^{-1}(\cos(\cos^{-1}x)\cos(\cos^{-1}(7x))-\sin(\cos^{-1}x)\sin(\cos^{-1}(7x)))$$
$$=\cos^{-1}(\cos(\cos^{-1}x+\cos^{-1}(7x)))=\cos^{-1}x+\cos^{-1}(7x).$$
  So 
  $$I:=\int \ln x \cos^{-1}(7x^2-\sqrt{49x^4-50x^2+1})\ \mathrm dx=\int \ln x \ (\cos^{-1}x+\cos^{-1}(7x))\ \mathrm dx.$$
  The rest of the solution is straightforward. Use integration by parts with $\ln x=u$ and $(\cos^{-1}x+\cos^{-1}(7x)) \ \mathrm dx=\mathrm dv.$ Then 
  $$\mathrm du=\frac{\mathrm dx}{x}, \ \ \ \ \ \ v=x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}$$
  and hence
  $$I=\ln x \left(x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}\right)-\int \left(\cos^{-1}x+\cos^{-1}(7x)-\frac{\sqrt{1-x^2}}{x}-\frac{\sqrt{1-49x^2}}{7x}\right)\ \mathrm dx$$
$$=(\ln x-1) \left(x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}\right)+\int \left(\frac{\sqrt{1-x^2}}{x}+\frac{\sqrt{1-49x^2}}{7x}\right)\mathrm dx. \ \ \ \ \ \ \ \ \ \ (1)$$
  Also, given $a > 0,$ the substitution $1-a^2x^2=t^2$ gives
  $$\int \frac{\sqrt{1-a^2x^2}}{x} \ \mathrm dx=\sqrt{1-a^2x^2}+\ln x-\ln(1+\sqrt{1-a^2x^2})+ \text{constant}. \ \ \ \ \ \ \ \ \ \ \ (2)$$
  Now $(1),(2)$ together complete the solution.

Brilliantly done; as always by him.
