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