Answer
$$\int x b^{x} d x =b^{x}\left(\frac{x}{\ln b}-\frac{1}{\ln ^{2} b}\right)+C$$
Work Step by Step
Given $$\int x b^{x} d x=b^{x}\left(\frac{x}{\ln b}-\frac{1}{\ln ^{2} b}\right)+C$$
Let
\begin{align*}
u&= x\ \ \ \ \ \ \ \ \ \ dv=\frac{1}{\ln b} b^xdx\\
du&=dx \ \ \ \ \ \ \ \ v=\frac{1}{\ln b} b^x
\end{align*}
Then
\begin{align*}
\int x b^{x} d x&=\frac{x}{\ln b} b^x-\int\frac{1}{\ln b} b^xdx \\
&=b^{x}\left(\frac{x}{\ln b}-\frac{1}{\ln ^{2} b}\right)+C
\end{align*}
