Answer
\begin{aligned}
\int x\ \ 5^{x}\ d x = \frac{\ 5^x}{\ln 5}(x- \frac{ 1}{\ln 5})+C \\
\end{aligned}
Work Step by Step
Given $$
\int x\ \ 5^{x}\ d x
$$
Use integration by parts:
$$ u= x \Rightarrow du= dx $$ $$ dv= 5^x \ dx \Rightarrow v= \frac{5^x}{\ln 5} $$ So, we get
\begin{aligned}
I&=\int x\ \ 5^{x}\ d x\\
&=uv - \int vdu\\
&= \frac{x\ 5^x}{\ln 5}-\int \frac{5^x}{\ln 5}dx\\
&= \frac{x\ 5^x}{\ln 5}- \frac{1}{\ln 5}\int 5^x dx\\
&= \frac{x\ 5^x}{\ln 5}- \frac{ 5^x}{(\ln 5)^2}+C \\
&= \frac{\ 5^x}{\ln 5}(x- \frac{ 1}{\ln 5})+C \\
\end{aligned}
