Difficult definite integration $\int_{-2}^{0}\frac{x^2 + x - 5}{(x-1)^2}e^x\,\mathrm dx$

Question

I found an integral in a contest that seems very difficult to compute. The answer is $-2$, however, I do not know how to arrive at this answer.

$$\int_{-2}^{0}\frac{x^2 + x - 5}{(x-1)^2}e^x\,\mathrm dx$$

At first I tried to make the substitution $u = x - 1$, but I did not get anywhere. I also tried to expand the denominator and perform synthetic division, which did not help so much either.

Also, I don't think it is possible to do partial fraction decomposition since the degree of the denominator is equal to the degree of the numerator.

---

Attempt with substitution (Moo's help) -

Let $u = x - 1$. Then, $du = 1$ $$\begin{align} I &= \int_{-2}^{0} \frac{x^2 + x - 5}{(x - 1)^2} e^x\,\mathrm dx =\\ &= \int_{-2}^{0} \frac{x^2 + x - 5}{u^2} e^{u+1}\,\mathrm du =\\ &= \int_{-2}^{0} \frac{e^{u+1}(x^2 + x - 5)}{u^2}\,\mathrm du \end{align}$$

Now, $x = u + 1$, $x^2 = u^2 + 2u + 2$ and therefore $$I = \int_{-3}^{-1} \frac{e^{u+1}(u^2 + 3u - 2)}{u^2}\mathrm du = e \int_{-3}^{-1} e^{u}\frac{u^2 + 3u - 3}{u^2}\mathrm du$$

Answer 1

Substituting for $u=x-1$, we get, $$I=\int_{-2}^{0} \frac {x^2-x+5}{(x-1)^2} e^x dx =\int_{-3}^{-1} \frac {u^2+3u-3}{u^2} e^{u+1} du $$ $$\Rightarrow I=e [\int_{-3}^{-1} [3 (\frac {e^u}{u} -\frac {e^u}{u^2}) du] +\int_{-3}^{-1} e^u du] =e [3I_1-3I_2 +I_3] $$ Now we have $$I_1-I_2 =\int_{-3}^{-1}(\frac {e^u}{u}-\frac {e^u}{u^2}) du =\int_{-3}^{-1} d (\frac {e^u}{u}) =\frac {e^u}{u} \mid_{-3}^{-1} $$ We see that $I_3$ is easy to solve. Finally the answer is $-2$. Hope it helps.

Answer 2

Integration by part we get \begin{align*} \int\frac{x^2 + x - 5}{(x-1)^2}e^x \, \mathrm{d}x&=-\frac{x^2 + x - 5}{x-1}e^x +\int e^{x}\left ( x+4 \right )\, \mathrm{d}x \\ &=-\frac{x^2 + x - 5}{x-1}e^x+e^{x}\left ( x+3 \right )\\ &=\frac{x+2}{x-1}e^{x} \end{align*} Then you can take it from here.