Answer:
b and c, b and c are same
Step-by-step explanation:
Only a and d can be evaluated using simple substitution.
For a, integral of the square root of (x-1)dx i.e
let y = x-1
dy = dx
It becomes integral square root of ydy, ∫√ydy
which is can easily be evaluated and y is substituted back.
For d, integral of x times the square root of the quantity x squared minus 1
let y= x squared minus 1
then dy = 2xdx
dx = dy/2x
It becomes integral x times square root of y dy/2x
∫x√ydy/2x=∫√ydy/2=1/2∫√ydy which can easily be evaluated and y is substituted back