lee-robinson_60b7316b3cdd1
lee-robinson_60b7316b3cdd1
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Aplicando una sustitución si ln x = u --> 1/x dx = du --> dx = x du --> dx = e^u du S Sin (ln x) dx = S sin u . E^u du aplicando integración por partes sin u = f(u) --> f ´(u) = cos u e^u = g ´(u) --> g(u) = e^u resulta S Sin (ln x) d…