<s>[INST] 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 $1, 2$
 [/INST] While $\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$ is only valid for $x\not\in\{1,\,2\}$ for the values of $A,\,B$ we seek, $2x+1=A(x-2)+B(x-1)$ will be valid for all $x\in\Bbb R$ because its validity at other values implies validity on $x\in\{1,\,2\}$ by continuity. You could also find $A,\,B$ from simultaneous equations obtained at other values of $x$, but using $x\in\{1,\,2\}$ is especially convenient, viz. $3=-A,\,5=B$. Or you could just equate coefficients of $x^1=x,\,x^0=1$, viz. $A+B=2,\,-2A-B=1$. That approach will serve you well for other kinds of coefficient-inferring problems.
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