Integral of exponential function

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$\int \left(5x-2\right)e^{5x^{2}-4x+2} \; dx$

**Answer**

$\int \left(5x-2\right)e^{5x^{2}-4x+2} \; dx$

Substitute $u= 5x^{2}-4x+2$. So, $du = \left(10x-4 \right) \; dx=2\left(5x-2\right) \;dx$

$\int \left(5x-2\right)e^{5x^{2}-4x+2} \; dx$

$\int \frac{1}{2}\left(10x-4\right)e^{5x^{2}-4x+2} \; dx$

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

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

Substituting back $u=5x^{2}-4x+2$

$=\frac{1}{2} e^{5x^{2}-4x+2} + C$