Answer
\[ - 3\ln \left| x \right| - 10{e^{ - 0.4}} + {e^{0.1}}x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( { - \frac{3}{x} + 4{e^{ - 0.4x}} + {e^{0.1}}} \right)dx} \hfill \\
By\,\,the\,\,sum\,\,and\,\,differentiate\,\,rule \hfill \\
- \int_{}^{} {\frac{3}{x}dx} + \int_{}^{} {4{e^{ - 0.4x}}dx} + \int_{}^{} {{e^{0.1}}\,dx} \hfill \\
Use\,\,indefinite\,\,integrals \hfill \\
\int_{}^{} {{x^{ - 1}}dx} = \ln \left| x \right| + C\,\,,\,\,\int_{}^{} {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C \hfill \\
Then \hfill \\
- 3\ln \left| x \right| + 4\,\left( {\frac{{{e^{ - 0.4x}}}}{{ - 0.4}}} \right) + {e^{0.1}}x + C \hfill \\
Simplifying \hfill \\
- 3\ln \left| x \right| + 4\,\left( { - \frac{5}{2}{e^{ - 0.4x}}} \right) + {e^{0.1}}x + C \hfill \\
- 3\ln \left| x \right| - 10{e^{ - 0.4}} + {e^{0.1}}x + C \hfill \\
\hfill \\
\end{gathered} \]
