For the given integral problem:` int_0^(ln(5))e^x/(1+e^(2x))dx` , it resembles the basic integration formula for inverse tangent:
`int_a^b (du)/(u^2+c^2) = (1/c)arctan(u/c) |_a^b`
where we let:
`u^2 =e^(2x) ` or` (e^x)^2 ` then `u= e^x`
`c^2 =1` or `1^2` then `c=1`
For the derivative of `u =e^(x)` , we apply the derivative of exponential function:
`du =e^x dx` .
Applying u-substitution: `u = e^x ` and` du = e^x dx` , we get:
`int e^x/(1+e^(2x))dx =int (e^xdx)/(1+(e^x)^2)`
`=int (du)/(1+(u)^2)`
Applying the basic integral formula of inverse tangent, we get:
`int (du)/(1+(u)^2) =(1/1)arctan(u/1)`
= `arctan(u)`
Express it in terms of x by plug-in `u=e^x` :
`arctan(u) =arctan(e^x)`
Evaluate with the given boundary limit:
`arctan(e^x)|_0^(ln(5)) =arctan(e^(ln(5)))-arctan(e^0)`
` =arctan(5)-arctan(1)`
` =arctan(5) -pi/4`
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