Họ nguyên hàm của hàm số sau \(f\left(x\right)=\dfrac{2^x-1}{e^x}\) là
\(\dfrac{2^x}{e^x}+e^{-x}+C\).\(\dfrac{2^x}{e^x\left(\ln2-1\right)}+e^{-x}+C\).\(\dfrac{2^x}{e^x\left(\ln2-1\right)}+C\).\(\dfrac{2^x}{\ln2-1}+e^{-x}+C\).Hướng dẫn giải:Ta có \(f\left(x\right)=\dfrac{2^x-1}{e^x}=\left(\dfrac{2}{e}\right)^x-e^{-x}\) nên
\(\int\left[\left(\dfrac{2}{e}\right)^x-e^{-x}\right]\text{dx}=\int\left(\dfrac{2}{e}\right)^x\text{dx}-\int e^{-x}\text{dx}\)\(=\dfrac{\left(\dfrac{2}{e}\right)^x}{\ln\left(\dfrac{2}{e}\right)}+\int e^{-x}\text{d}\left(-x\right)\)\(=\dfrac{2^x}{e^x\left(\ln2-1\right)}+e^{-x}+C\) .