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,