Finding $\int \frac{e^x\left(-2x^2+12x-20\right)}{x^3-6x^2+12x-8}dx$

Question

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.

Answer 1

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.

Answer 2

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

$$\frac{dt}{t} = 1 - \frac{1}{3-x} + \frac{2}{2-x}$$ $$\implies dt = (1 + \frac{1}{3-x} + \frac{2}{2-x})t$$ $$\implies dt = \frac{e^x(x^2 + 12x - 20)}{(x-2)^3}$$

Answer 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.