Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator to get $$2x+1=A(x-2)+B(x-1)$$ Now to find $A$ and $B$ how can we put $x=1$ and $x=2$ in this identity as this identity is valid if and only if $x$ is not equal to $1, 2$

The expression
(2x+1)/(x-1)(x-2). = A / ( x-1). + B/(x-2) is discontinuous at x= 1 and at x= 2 but identy for restall value of x.
Once we write as 2x +1 = A (x-2) + B (x-1) since left hand side is continuous thus Right-hand side will be having same behaviour .hence we can put x= 2 and x= 1 to find A and B
