#\int_0^1e^x/(1+e^(2x))dx=?#

Question

please don't use LaGrange (#\lambda#), I'm only in Calculus II

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Answer 1 · 2017-09-15T20:42:34

#arctan(e)-pi/4#

Let #u=e^x#. Thus #du=e^xdx#. The bounds are also transformed: #x=0=>u=e^0=1# and #x=1=>u=e^1=e#.

#int_0^1e^x/(1+e^(2x))dx=int_1^e1/(1+u^2)du#

This is the arctangent integral:

#=arctan(u)|_1^e=arctan(e)-arctan(1)=arctan(e)-pi/4#

#approx0.43288474#