Answer
$\displaystyle (1-x^2)\frac{2^{-x}}{\ln 2}+\frac{2^{1-x}x}{(\ln 2)^2}-\frac{2^{1-x}}{(\ln 2)^3}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int(1-x^2)2^{-x} \ dx=\displaystyle (1-x^2)(\frac{-2^{-x}}{\ln 2})- \int (-\frac{2^{-x}}{\ln 2})(-2x) \ dx$
or, $=\displaystyle (1-x^2)\frac{2^{-x}}{\ln 2}+\frac{2^{1-x}}{(\ln 2)^2}+\dfrac{2}{(\ln 2)^2}\int 2^{-x} \ dx$
or, $=\displaystyle (1-x^2)\frac{2^{-x}}{\ln 2}+\frac{2^{1-x}x}{(\ln 2)^2}-\frac{2^{1-x}}{(\ln 2)^3}+C$
