Integral of exponential function

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$\int xe^{9x^{2}} \; dx$

**Answer**

$\int xe^{9x^{2}} \; dx$

Substitute $u= 9x^{2}$. So, $du = 18x\; dx$, or $x\;dx =\frac{1}{18}\; du $

Then $\int x e^{9x^{2}} \; dx$

$=\frac{1}{18}\int e^u \; du $

$=\frac{1}{18}e^u + C $

Substituting back $u=9x^{2}$

$=\frac{1}{18} e^{9x^{2}} + C$