Answer
\[ = - \frac{1}{4}{e^{\,\left( {3 - 4x} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{e^{\,\left( {3 - 4x} \right)}}\,dx} \hfill \\
\hfill \\
substituting\,\,\,\left( {3 - 4x} \right) = u\,\,and\,\,dx = - \frac{{du}}{4} \hfill \\
\hfill \\
= - \frac{1}{4}\int_{}^{} {{e^u}du} \hfill \\
\hfill \\
integrate\,\,\,{\text{using }}\int {{e^u}du = {e^u} + C} \hfill \\
\hfill \\
= - \frac{1}{4}{e^u} + C \hfill \\
\hfill \\
substituting\,back\,\,\,u = 3 - 4x \hfill \\
\hfill \\
= - \frac{1}{4}{e^{\,\left( {3 - 4x} \right)}} + C \hfill \\
\end{gathered} \]
