Integral of exponential function

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$\int \left(6x+1\right)e^{3x^{2}+x-9} \; dx$

**Answer**

$\int \left(6x+1\right)e^{3x^{2}+x-9} \; dx$

Substitute $u= 3x^{2}+x-9$. So, $du = \left(6x+1 \right) \; dx$

$\int \left(6x+1\right)e^{3x^{2}+x-9} \; dx$

$=\int e^u \; du $

$=e^u + C $

Substituting back $u=3x^{2}+x-9$

$= e^{3x^{2}+x-9} + C$