How do you find the integral ∫1x2⋅√x2−9dx ?
1 Answer
Aug 29, 2014
=19x⋅√x2−9+c , wherec is a constantExplanation :
=∫1x2√x2−9dx
=∫1x3√1−9x2dx Using Integration by Substitution,
let's assume
3x=t ,−3x2dx=dt
=∫−t9√1−t2dt=−19∫t√1−t2dt again assuming
t2=z ,2tdt=dz
=−19∫12⋅dz√1−z
=−118∫1√1−zdz
=−118∫(1−z)−12dz
=118⋅(1−z)1212+c , wherec is a constantSubstituting value of
z andt back,
=19√1−9x2+c , wherec is a constant
=19x⋅√x2−9+c , wherec is a constant
