Integral of exponential function

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$\int \left(2x-3\right)e^{x^{2}-3x-3} \; dx$

**Answer**

$\int \left(2x-3\right)e^{x^{2}-3x-3} \; dx$

Substitute $u= x^{2}-3x-3$. So, $du = \left(2x-3 \right) \; dx$

$\int \left(2x-3\right)e^{x^{2}-3x-3} \; dx$

$=\int e^u \; du $

$=e^u + C $

Substituting back $u=x^{2}-3x-3$

$= e^{x^{2}-3x-3} + C$