Evaluate:


Let


Put t=x2


dt=2xdx


Now putting,


A(t+3) +B(t+1) = 1


Putting t+3=0,


X=-3


A(0) +B(-3+1)=1



Putting t+1=0,


X=-1


A(-1+3)+B(0)=1



From equation(1),we get,






1