How do you find the integral int1/(x^2*sqrt(x^2-9))dx∫1x2⋅√x2−9dx ?
1 Answer
Aug 29, 2014
=1/(9x)*sqrt(x^2-9)+c=19x⋅√x2−9+c , wherecc is a constantExplanation :
=int1/(x^2sqrt(x^2-9))dx=∫1x2√x2−9dx
=int1/(x^3sqrt(1-9/x^2))dx=∫1x3√1−9x2dx Using Integration by Substitution,
let's assume
3/x=t3x=t ,-3/x^2dx=dt−3x2dx=dt
=int-t/(9sqrt(1-t^2))dt=-1/9intt/(sqrt(1-t^2))dt=∫−t9√1−t2dt=−19∫t√1−t2dt again assuming
t^2=zt2=z ,2tdt=dz2tdt=dz
=-1/9int1/2*dz/sqrt(1-z)=−19∫12⋅dz√1−z
=-1/18int1/sqrt(1-z)dz=−118∫1√1−zdz
=-1/18int(1-z)^(-1/2)dz=−118∫(1−z)−12dz
=1/18*(1-z)^(1/2)/(1/2)+c=118⋅(1−z)1212+c , wherecc is a constantSubstituting value of
zz andtt back,
=1/9sqrt(1-9/x^2)+c=19√1−9x2+c , wherecc is a constant
=1/(9x)*sqrt(x^2-9)+c=19x⋅√x2−9+c , wherecc is a constant
