This can be rewritten as intcos^-1x x^-2dx. Whenever you see a product within an integral, you should consider using integration by parts.
The integration by parts formula states that an integral intudv = uv - int(v du). cos^-1x is not that simple to integrate, so let's say u = cos^-1x and dv= x^-2. Then du = -1/sqrt(1 - x^2)dx and v = -1/x.
Use the formula now:
intcos^-1x x^-2dx = -cos^-1x/x - int(-1/x * -1/sqrt(1 - x^2)dx)
intcos^-1x x^-2dx = -cos^-1x/x - int 1/(xsqrt(1 - x^2)) dx
The remaining integral will have to be integrated by trigonometric substitution. This is a technique of integration where you make use of a pythagorean identity to get rid of the √. When the square root within the integral is of the form sqrt(a^2 - x^2) like the one above, use the substitution color(red)(x = sin theta). Then dx = costheta d theta.
int1/(xsqrt(1 - x^2)) dx= int1/(sinthetasqrt(1 - sin^2theta)) * costheta d theta
int1/(xsqrt(1 - x^2))dx =int 1/(sin thetasqrt(cos^2theta)) *costheta d theta
int1/(xsqrt(1 - x^2))dx = int1/sintheta d theta
int 1/(xsqrt(1 - x^2))dx = intcsctheta d theta
This is a known integral that is derived here.
int 1/(xsqrt(1 - x^2))dx = -ln|csctheta + cottheta| + C
We now reverse the substitutions by drawing a triangle and finding the correct ratios. We know from our original substitution that x/1 = sintheta.
int 1/(xsqrt(1 - x^2))dx = -ln|1/x + sqrt(1 - x^2)/x| + C
int 1/(xsqrt(1 - x^2))dx = -ln|(1 + sqrt(1 - x^2))/x| + C
Let's return our attention to the whole integral.
intcos^-1x x^-2dx = -cos^-1x/x- (- ln|(1 + sqrt(1 - x^2))/x|) + C
intcos^-1x x^-2dx = ln|(1 + sqrt(1 - x^2))/x| - cos^-1x/x + C
Hopefully this helps!
