\(F\left(x\right)\) là một nguyên hàm của \(f\left(x\right)=\dfrac{2^x-1}{e^x}\) và \(F\left(0\right)=1\). Khẳng định nào sau đây đúng?
\(F\left(x\right)=\dfrac{2^x+\ln2-1}{e^x\left(\ln2-1\right)}\). \(F\left(x\right)=\dfrac{1}{\ln2-1}\left(\dfrac{2}{e}\right)^x+\left(\dfrac{1}{e}\right)^x-\dfrac{1}{\ln2-1}\). \(F\left(x\right)=\dfrac{2^x+\ln2}{e^x\left(\ln2-1\right)}\). \(F\left(x\right)=\left(\dfrac{2}{e}\right)^x\). Hướng dẫn giải:\(F\left(x\right)=\int\dfrac{2^x-1}{e^x}\text{d}x=\int\left[\left(\dfrac{2}{e}\right)^x-e^{-x}\right]\text{d}x=\int\left(\dfrac{2}{e}\right)^x\text{d}x+\int e^{-x}\text{d}\left(-x\right)\)
\(=\dfrac{\left(\dfrac{2}{e}\right)^x}{\ln\left(\dfrac{2}{e}\right)}+e^{-x}+C\)
\(=\dfrac{1}{\ln2-1}\left(\dfrac{2}{e}\right)^x+\left(\dfrac{1}{e}\right)^x+C\)
Để \(F\left(0\right)=1\) thì
\(\dfrac{1}{\ln2-1}\left(\dfrac{2}{e}\right)^0+\left(\dfrac{1}{e}\right)^0+C=1\)
\(\Rightarrow C=-\dfrac{1}{\ln2-1}\)
\(\Rightarrow F\left(x\right)=\dfrac{1}{\ln2-1}\left(\dfrac{2}{e}\right)^x+\left(\dfrac{1}{e}\right)^x-\dfrac{1}{\ln2-1}\).