Answer
$\frac{x^3}{3}-\frac{1}{2^x \ln 2}+C$
Work Step by Step
Let $\log x$ denote the natural logarithm $\ln x$.
$\int(x^2)+2^{-x}$ -> $\int x^2$ $dx$ $+$ $\int2^{-x}$ $dx$
let $-x=u$
$-dx=dx$
$\int2^{-x}$ $dx$
$=-\int 2^u$ $du$
$=-\frac{2^u}{\log 2}$
$=-\frac{1}{2^x \log 2}$
Therefore,
$\int(x^2)+2^{-x}$
$=\frac{x^3}{3}-\frac{1}{2^x \log 2}+C$
