How to prove that $$\int \dfrac{e^x\left(-2x^2+12x-20\right)}{(x-2)^3}dx=-\frac{2e^x(x-3)}{(x-2)^2}$$ without using the quotient rule for derivatives $\left(\frac{f}{g}\right)'=\frac{gf'-g'f}{g^2}$ (suppose we do not know the solution)
Any hint would be appreciated.
Let's ignore the factor of $-2$ for now in the numerator, since we can adjust it later by multiplying both sides by $-2$.
Taking $t$ as $x-2$ in the L.H.S we get the integral $$\int \frac{e^{t+2}(t^{2} -2t +2)}{t^{3}}dt$$
Applying ILATE rule on $\int\frac{e^{t+2}t^2}{t^3}dt$ we get,
$$ = \frac{e^{t+2}}{t} + \int \frac{e^{t+2}}{t^{2}}dt - 2\int \frac{e^{t+2}}{t^2}dt + 2 \int \frac{e^{t+2}}{t^3}dt$$
$$ = \frac{e^{t+2}}{t} - \int \frac{e^{t+2}}{t^2}dt + 2 \int \frac{e^{t+2}}{t^3}dt$$
Using the ILATE rule once again on the middle term in the above expression, we get
$$ = \frac{e^{t+2}}{t} - \left(\frac{e^{t+2}}{t^2} + 2\int \frac{e^{t+2}}{t^3}dt\right) + 2 \int \frac{e^{t+2}}{t^3}dt$$
So the answer is
$$ = \frac{e^{t+2}}{t} - \frac{e^{t+2}}{t^2}$$
Now replace $t = x-2$ in the above equation
$$\frac{e^{x}(x-3)}{(x-2)^2}$$
Now just multiply LHS and RHS by $-2$ to get the desired form.
Q.E.D.
If we already know the RHS (and if you don't want to use the quotient rule) you can simply do this -
Let $$t = - \frac{2e^x(x-3)}{(x-2)^2}$$
Taking $\log$ on both sides of the above equation, we get.
$$\log{t} = \log 2 + x + \log(3-x) - 2\log(x-2)$$
Differentiating the above equation on both sides we get
$$\frac{dt}{t} = 1 - \frac{1}{3-x} + \frac{2}{2-x}$$ $$\implies dt = (1 + \frac{1}{3-x} + \frac{2}{2-x})t$$ $$\implies dt = \frac{e^x(x^2 + 12x - 20)}{(x-2)^3}$$
Because the denominatof the integrand is of degree $3$, let us assume that the integral is $$\frac{e^x P_n(x)}{(x-2)^2}$$ in which $P_n(x)$ is a polynomial of degree $n$.
Now compute the derivative $$\frac d {dx}\Big(\frac{e^x P_n(x)}{(x-2)^2}\Big)=\frac{e^x \left((x-2) P'(x)+(x-4) P(x)\right)}{(x-2)^3}$$ So, the numerator is of degree $n+1$ that is to say that we need to consider $n=1$. So, write $P_1(x)=a+bx$ and replace; this gives $$(x-2) P'(x)+(x-4) P(x)=-(4 a+2 b)+x (a-3 b)+b x^2$$ and this must be equal to $-2x^2+12x-20$.
Compare the coefficients.
The integral is of the form \begin{equation*} \int \left( \frac{-2x^{2}+12x-20}{(x-2)^{3}}\right) e^{x}dx=\int h(x)e^{x}dx. \end{equation*} This form recalls the well-known formula \begin{equation*} \int \left( f^{\prime }(x)+f(x)\right) e^{x}dx=f(x)e^{x}+C. \end{equation*} The proof of a more general formula maybe found at
Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$
So we are done if we find a function $f(x)$ such that \begin{equation*} h(x)=\frac{-2x^{2}+12x-20}{(x-2)^{3}}=f^{\prime }(x)+f(x). \end{equation*} In what follows, I will show that $f(x)=\frac{-2(x-3)}{(x-2)^{2}}$ and therefore \begin{equation*} \int \left( \frac{-2x^{2}+12x-20}{(x-2)^{3}}\right) e^{x}dx=\left( f(x)\right) e^{x}+C=\frac{-2(x-3)}{(x-2)^{2}}e^{x}+C. \end{equation*} $\color{red}{\bf Problem:}$ We want to write $\frac{-2x^{2}+12x-20}{(x-2)^{3}}$ as $ f^{\prime }(x)+f(x)$ and $f(x)$ is to be determined.
First, divide $-2x^{2}+12x-20$ by $(x-2)^{2}.$ We get \begin{equation*} -2x^{2}+12x-20=-2(x-2)^{2}+4x-12 \end{equation*} then \begin{equation*} \frac{-2x^{2}+12x-20}{(x-2)^{3}}=\frac{-2(x-2)^{2}+4x-12}{(x-2)^{3}}=\frac{-2 }{x-2}+\frac{4x-12}{(x-2)^{3}}. \end{equation*} Now, let \begin{equation*} f_{1}(x)=\frac{-2}{x-2},\ \ \ \ \ \ \ then\ \ \ \ \ \ \ \ f_{1}^{\prime }(x)= \frac{2}{(x-2)^{2}} \end{equation*} and add and subtract it derivative \begin{equation*} \frac{-2x^{2}+12x-20}{(x-2)^{3}}=\left( f_{1}(x)+f_{1}^{\prime }(x)\right) - \frac{2}{(x-2)^{2}}+\frac{4x-12}{(x-2)^{3}}=\left( f_{1}(x)+f_{1}^{\prime }(x)\right) +\frac{2x-8}{(x-2)^{3}} \end{equation*} Now repeat the same procedure but with the new and reduced fraction $\frac{2x-8}{ (x-2)^{3}}.$ So, divide $2x-8$ by $(x-2)$ to obtain $2x-8=2(x-2)-4$, then \begin{equation*} \frac{2x-8}{(x-2)^{3}}=\frac{2(x-2)-4}{(x-2)^{3}}=\frac{2}{(x-2)^{2}}-\frac{4 }{(x-2)^{3}}. \end{equation*} Now let \begin{equation*} f_{2}(x)=\frac{2}{(x-2)^{2}}, \end{equation*} and add and subtract its derivative $f_{2}^{\prime }(x)=\frac{-4}{(x-2)^{3}}.$ \begin{equation*} \frac{2x-8}{(x-2)^{3}}=\left( f_{2}(x)+f_{2}^{\prime }(x)\right) -f_{2}^{\prime }(x)-\frac{4}{(x-2)^{3}}=\left( f_{2}(x)+f_{2}^{\prime }(x)\right) \end{equation*} Therefore \begin{equation*} \frac{-2x^{2}+12x-20}{(x-2)^{3}}=\left( f_{1}(x)+f_{1}^{\prime }(x)\right) +\left( f_{2}(x)+f_{2}^{\prime }(x)\right) =(f_{1}(x)+f_{2}(x))+(f_{1}(x)+f_{2}(x))^{\prime } \end{equation*} It suffices to take \begin{equation*} f(x)=f_{1}(x)+f_{2}(x)=\frac{-2}{x-2}+\frac{2}{(x-2)^{2}}=\frac{-2(x-3)}{ (x-2)^{2}}. \end{equation*} The integral is
\begin{equation*} \int \left( \frac{-2x^{2}+12x-20}{(x-2)^{3}}\right) e^{x}dx=\left( f(x)\right) e^{x}+C=\frac{-2(x-3)}{(x-2)^{2}}e^{x}+C. \ \ \color{red} \blacksquare \end{equation*}
