Chọn phương án A.

Đặt \(\begin{cases}
u=x\\ v'=2^x
\end{cases}\Rightarrow\begin{cases}
u'=1\\ v=\dfrac{2^x}{\ln2}
\end{cases}\)

Khi đó $$\begin{aligned}I=&\dfrac{x\cdot2^x}{\ln2}\bigg|_0^1-\displaystyle\int\limits_{0}^{1}\dfrac{2^x}{\ln2}\mathrm{\,d}x\\ =&\dfrac{2}{\ln2}-\dfrac{2^x}{\ln^22}\bigg|_0^1=\dfrac{2\ln2-1}{\ln^22}.\end{aligned}$$