Answer
$-e^{-(x^2-3x+1)}+C=-\dfrac{1}{e^{(x^2-3x+1)}}+C$
Work Step by Step
We will solve the given integral by using u-substitution method.
Let us consider that $u=x^2-3x+1 \implies du=(2x-3) \ dx$
$\displaystyle \int \dfrac{2x-3}{(e^{x^2-3x+1})} \ dx=\int \displaystyle \dfrac{du}{e^u} $
or, $=\int e^{-u} \ du$
or, $=-e^{-u}+C$
Now, we will use back substitution $u=x^2-3x+1$
Therefore, we have:
$\displaystyle \int \dfrac{2x-3}{(e^{x^2-3x+1})} \ dx=-e^{-(x^2-3x+1)}+C=-\dfrac{1}{e^{(x^2-3x+1)}}+C$
