Question

Using second fundamental theorem, evaluate the following:

${\displaystyle {\int }_0^3\frac{{\text{e}}^x\text{d}x}{1+{\text{e}}^x}}$

Answer 1

${\displaystyle {\int }_0^3\frac{{\text{e}}^x\text{d}x}{1+{\text{e}}^x}={\left\{\log |1+\text{e}x|\right\}}_0^3}$

= log |1 + e³| – log |1 + e°|

= log |1 + e³| – log |1 + 1|

= log |1 + e³| – log |2|

= ${\displaystyle \log \left|\frac{1+{\text{e}}^3}{2}\right|}$