Integrate $\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$ I need integrate:
$$\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$$
How can i solve it? is it good way substitute argument of arccos? $$t=\sqrt{\frac{x-4}{x+6}}$$ 
 A: Hint. Here is a route.
By the change of variable 
$$
t=\sqrt{\dfrac{x-4}{x+6}}\qquad x= 2\:\dfrac{ 3 t^2+2}{1-t^2} \qquad dx= 2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'dt
$$ one may obtain
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\int\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'\arccos(t)\:dt
$$ then one may integrate by parts to get
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)\arccos(t)+2\int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt
$$ The latter integral may be evaluated by the change of variable $t:=\sin u$ yielding
$$
\int\dfrac{ 3 t^2+2}{(1-t^2)^{3/2}}\:dt=\frac{10 t}{\sqrt{1-t^2}}-6\: \text{arcsin}(t)+C, \quad |t|<1.
$$
A: Here is another route:
First integrating by parts, writing the integral as
$$
\int 1\cdot \arccos\sqrt{\frac{x-4}{x+6}}\,dx
$$
will leave you with an integral like (I forget about constants, and suggest you to do the calculations!)
$$
\int \frac{x}{\sqrt{x-4}(x+6)}\,dx.
$$
Adding and subtracting $6$ in the numerator (and integrating the part looking like $1/\sqrt{x-4}$) will leave you with something like
$$
\int\frac{1}{\sqrt{x-4}(x+6)}\,dx
$$
Now, let $t=\sqrt{x-4}$ and you will get something like (again, check constants)
$$
\int\frac{1}{t^2+10}\,dt,
$$
which will just give an arctan. 
If you want to know what I actually got in the end, move your mouse over the box below.

 $$ x \arccos\sqrt{\frac{x-4}{x+6}}+\sqrt{10(x-4)}-6\arctan\sqrt{\frac{x-4}{10}}+C$$

