For which values of $a,b$ the following integral is an elementary function, and which elementary function?
$$\int \frac{x^2+ax+b}{(x-1)^2}\,e^x\, dx$$
I tried to solve this integral but it is too hard, someone can help me please, thanks for your time and help.
On
it is very unlikely that any function multiplied by $e^x$ has an elemental derivative, so the goal is to make $\frac{x^2+ax+b}{(x-1)^2}$ into a constant.
expand the bottom of the fraction. $$\frac{x^2+ax+b}{x^2-2x+1}$$
As you can see, the top and bottom line up perfectly. if $a=-2$ and $b=1$, the numerator and denominator become the same, so your function simplifies into $\int{e^x}dx$, and is elemental.
I don't know if there are more correct answers, but it is safe to assume that there are not any, because any other values for $a$ and $b$ would result in the integral becoming more complicated than is expected of you in precalc.
tl/dr: a=-2, b=1.
$$\frac{d}{dx}\frac{e^x}{(x-1)} = \frac{(x-2)}{(x-1)^2}e^{x} $$ hence if $$ \frac{x^2+ax+b}{(x-1)^2} = 1+k\frac{x-2}{(x-1)^2} $$ i.e. $\color{red}{2a+b = -3}$, we have an elementary integral, $\color{red}{\left(1+\frac{a+2}{x-1}\right)e^x}$.
Some differential Galois theory is required to prove that the previous if is in fact an iff.