Answer
$$\frac{1}{3}e^{3 x}-\frac{\left( 5 \right)^{-x}}{\ln 5}+x\ln (1.4x)- x+c$$
Work Step by Step
Given $$
\int\left[e^{3 x}+\ln (1.4 x)+\left(\frac{1}{5}\right)^{x}\right] d x
$$
Since
\begin{align*}
\int\left[e^{3 x}+\ln (1.4 x)+\left(\frac{1}{5}\right)^{x}\right] d x&=
\int\left[e^{3 x}+\ln (1.4 x)+\left( 5 \right)^{-x}\right] d x\\
&=\frac{1}{3}e^{3 x}-\frac{\left( 5 \right)^{-x}}{\ln 5}+\int\ln (1.4 x)d x
\end{align*}
To evaluate $ \int\ln (1.4 x)d x$, let $u=1.4x\ \to\ du=1.4dx$ and apply the rule $$
\int \ln u d u=u \ln u-u+c
$$
We get
\begin{align*}
\ln (1.4x)dx&=\frac{1}{1.4}\int\ln udu\\
&=\frac{1}{1.4}\left(u\ln u-u\right)+c\\
&=\frac{1}{1.4}\left(1.4x\ln (1.4x)-1.4x\right)+c\\
&=x\ln (1.4x)- x+c
\end{align*}
Hence
$$\int\left[e^{3 x}+\ln (1.4 x)+\left(\frac{1}{5}\right)^{x}\right] d x =\frac{1}{3}e^{3 x}-\frac{\left( 5 \right)^{-x}}{\ln 5}+x\ln (1.4x)- x+c$$
