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 $1, 2$

\begin{align*}
\frac{2x+1}{(x-1)(x-2)} & = \frac{(2x-2) + 3}{(x-1)(x-2)} = \frac{2(x-1)}{(x-1)(x-2)} + \frac{3}{(x-1)(x-2)}\\\\
& = \frac{2}{x-2} + 3\times\frac{(x-1) - (x-2)}{(x-1)(x-2)} = \frac{2}{x-2} + \frac{3}{x-2} - \frac{3}{x-1}\\\\
& = \frac{5}{x-2} - \frac{3}{x-1}
\end{align*}
