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 , wherec 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 , wherec is a constantSubstituting value of
z andt back,
=1/9sqrt(1-9/x^2)+c , wherec is a constant
=1/(9x)*sqrt(x^2-9)+c , wherec is a constant