Integral of exponential function

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$\int e^{-19-6x} \; dx$

**Answer**

$\int e^{-19-6x} \; dx$

Substitute $u= -19-6x$. So, $du = -6\; dx$, or $dx =-\frac{1}{6}\; du $

Then $\int e^{-19-6x} \; dx$

$=-\frac{1}{6}\int e^u \; du $

$=-\frac{1}{6}e^u + C $

Substituting back $u=-19-6x$

$=-\frac{1}{6} e^{-19-6x} + C$