Answer
$\displaystyle 2^x [\frac{2-x}{\ln 2}+\frac{1}{(\ln 2)^2}]+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int(2-x)2^x \ dx=\displaystyle (2-x)\frac{2^x}{\ln 2}-\int \frac{2^x}{\ln 2}(-1) \ dx$
or, $=\displaystyle (2-x)\frac{2^x}{\ln 2}+\dfrac{1}{\ln 2}\int 2^x \ dx$
or, $=\displaystyle (2-x)\frac{2^x}{\ln 2}+ \frac{2^x}{(\ln 2)^2}+C$
or, $=\displaystyle 2^x [\frac{2-x}{\ln 2}+\frac{1}{(\ln 2)^2}]+C$
