[tex]\\ \pi \int^1_0 (\frac{e^x}{e^x+1}) dx = \pi (ln|e^x+1||^1_0) = \pi (ln(e^x+1)|^1_0) \\ \\ \boxed{Schimbare \:\ variabila: u = e^x ; du = (e^x+1)` = e^x} \\ \\ \pi [ln(e^1+1) - ln(e^0+1)] =\ \textgreater \ \pi [ln(e+1) - ln(1+1)] \\ \\ \\ = \pi [ln(e+1) -ln(2)] \ \textless \ =\ \textgreater \ \boxed{\pi \:\ ln(\frac{e+1}{2})}[/tex]
