How do you find the integral int1/(x^2*sqrt(x^2-9))dx ?

1 Answer
Aug 29, 2014

=1/(9x)*sqrt(x^2-9)+c, where c is a constant

Explanation :

=int1/(x^2sqrt(x^2-9))dx

=int1/(x^3sqrt(1-9/x^2))dx

Using Integration by Substitution,

let's assume 3/x=t, -3/x^2dx=dt

=int-t/(9sqrt(1-t^2))dt=-1/9intt/(sqrt(1-t^2))dt

again assuming t^2=z, 2tdt=dz

=-1/9int1/2*dz/sqrt(1-z)

=-1/18int1/sqrt(1-z)dz

=-1/18int(1-z)^(-1/2)dz

=1/18*(1-z)^(1/2)/(1/2)+c, where c is a constant

Substituting value of z and t back,

=1/9sqrt(1-9/x^2)+c, where c is a constant

=1/(9x)*sqrt(x^2-9)+c, where c is a constant