This one is more obvious than your last one
(e^-x - 1)/(e^-x + x)
notice that the derivative of (e^-x + x) is
(-1)e^-x + 1
= -1(e^-x - 1)
you have exactly the opposite of that sitting in your numerator.
Which , by observation alone, suggest a log integration
since for y' = u'/u , y = ln u
so ∫(e^-x - 1)/(e^-x + x) dx
= - ln (e^-x + x) + c
find indefinite integral
((e^-x)-(1))/((e^-x)+(x))
1 answer
1 answer