Integral of exponential function

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$\int e^{-9x} \; dx$

**Answer**

$\int e^{-9x} \; dx$

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

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

$=-\frac{1}{9}\int e^u \; du $

$=-\frac{1}{9}e^u + C $

Substituting back $u=-9x$

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