Đáp án:
`\int_{0}^{ln2}\frac{e^{2x+1}+1}{e^x}dx=e+1/2`
Giải thích các bước giải:
`\int_{0}^{ln2}\frac{e^{2x+1}+1}{e^x}dx`
`=\int_{0}^{ln2}(\frac{e^{2x+1}}{e^x}+\frac{1}{e^x})dx`
`=\int_{0}^{ln2}(e^(x+1)+e^-x)dx`
`=(e^(x+1)-e^-x)`$\Bigg|_0^{\ln2}$
`=e^(ln2+1)-e^-ln2-e+e^0`
`=e+1/2`