Evaluate the integral? : # int 1/( x^2sqrt(x^2-4) ) dx #

We seek:

# I = int \ 1/( x^2sqrt(x^2-4) ) \ dx #

Because of the negative sign let us attempt the following substitution:

# x=2sec theta => x^2 = 4sec^2 theta #

And differentiating wrt #theta#, we get:

# (dx)/(d theta) =2sec theta tan theta #

Substituting into the integral we get:

# I = int \ 1/( (4sec^2theta)sqrt(4sec^2theta - 4) ) \ 2sec theta tan theta \ d theta #

# \ \ = 1/2 \ int \ 1/( secthetasqrt(4(sec^2theta - 1)) ) \ tan theta \ d theta #

# \ \ = 1/2 \ int \ 1/( secthetasqrt(4tan^2 theta) ) \ tan theta \ d theta #

# \ \ = 1/2 \ int \ 1/( sectheta(2tan theta) ) \ tan theta \ d theta #

# \ \ = 1/4 \ int \ 1/( sectheta ) \ d theta #

# \ \ = 1/4 \ int \ costheta \ d theta #

# \ \ = 1/4 \ sin theta + C # ..... [A]

We now have a simple solution, but we need to be able to restore the substitution; so we need to perform some further trigonometric manipulation. now:

# x=2sec theta => sec theta = x/2 #
# :. 1/(cos theta) = x/2 #
# :. cos theta = 2/x #

And using the pythagorean identity #sin^2A + cos^2A -= 1# then:

# sin^2theta + (2/x)^2 = 1 #
# :. sin^2theta = 1 - 4/x^2 = (x^2-4)/x^2 #
# :. sin theta = sqrt((x^2-4)/x^2) = sqrt(x^2-4)/x#

And now using this result we can restore the earlier substitution in [A], to give:

# I = 1/4 \ sqrt(x^2-4)/x + C #
# \ \ = sqrt(x^2-4)/(4x) + C #