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 constantExplanation :
#=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 constantSubstituting 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
