Integrating e^x/(e^{2x} + 1): Long Division?

AI Thread Summary
The discussion focuses on the integral ∫(e^x/(e^{2x} + 1) dx and the challenges of rewriting it to fit the arctan formula. Participants suggest that long division may not be necessary, emphasizing the effectiveness of u-substitution, specifically using u = e^x. The conversation highlights the importance of u-substitutions in simplifying integrals and solving problems efficiently. Overall, the consensus is to embrace u-substitution as a valuable tool in integration. Understanding these techniques can significantly ease the process of solving complex integrals.
RadiationX
Messages
255
Reaction score
0
i'm having trouble rewriting this integral:\int\frac{e^x}{e^{2x} + 1} so that it will be in the arctan formula: should i use long divison here? if it were not for the e^x in the numerator i'd be fine.
 
Last edited:
Physics news on Phys.org
u = e^{x}, du = e^{x}dx

e^{2x} = (e^{x})^2
 
you never get away from u -substitutions do ! thanks
 
Why woudl you watn to get away from them? They save your ass a lot :)
Learn to love em.
 
yeah your right
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top