Evaluate the integrals using substitution.
Given: 
Let x2 + 1 = t
⇒ 2xdx = dt
⇒ xdx = � dt
When x = 0, t = 1 and when x = 1, t = 2
1
Evaluate the integrals using substitution.
Given:
Let x2 + 1 = t
⇒ 2xdx = dt
⇒ xdx = � dt
When x = 0, t = 1 and when x = 1, t = 2