Integral of exponential function

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$\int \left(8x+3\right)e^{8x^{2}+6x-5} \; dx$

**Answer**

$\int \left(8x+3\right)e^{8x^{2}+6x-5} \; dx$

Substitute $u= 8x^{2}+6x-5$. So, $du = \left(16x+6 \right) \; dx=2\left(8x+3\right) \;dx$

$\int \left(8x+3\right)e^{8x^{2}+6x-5} \; dx$

$\int \frac{1}{2}\left(16x+6\right)e^{8x^{2}+6x-5} \; dx$

$=\frac{1}{2}\int e^u \; du $

$=\frac{1}{2}e^u + C $

Substituting back $u=8x^{2}+6x-5$

$=\frac{1}{2} e^{8x^{2}+6x-5} + C$