an interesting integral

[@avielabc604]

Using kings rule would solve it a lot quicker 🙌

[@emerald_eyes]

Just take divide numerator and denominator by 5^x. You get integral of dx/5^x(1+1/5^x)
Then you set u=1/5^x, du=ln(1/5)/5^x. dx
Substitute into the equation 1/ln(1/5) × integral of du/1+u. And the answer is ln(1+1/5^x)/ln(1/5)

[@seegeeaye]

divede 5^x is to multiply 5^-x, the same

[@ahmedsabri-ng2xx]

an interesting integral

[@noteventrying27]

mhm

[@gelbkehlchen]

Solution:
∫1/(5^x+1)*dx = ∫(5^x+1-5^x)/(5^x+1)*dx
= ∫(5^x+1)/(5^x+1)*dx-∫5^x/(5^x+1)*dx
= ∫dx-∫5^x/(5^x+1)*dx
= x-∫e^[x*ln(5)]/{e^[x*ln(5)]+1}*dx
= x-1/ln(5)*∫e^[x*ln(5)]/{e^[x*ln(5)]+1}*ln(5)*dx
---------------------
Substitution: u = e^[x*ln(5)]+1 du = ln(5)*e^[x*ln(5)]
---------------------
= x-1/ln(5)*∫1/u*du = x-1/ln(5)*ln|u|+C
= x-1/ln(5)*ln|e^[x*ln(5)]+1|+C
= x-1/ln(5)*ln|5^x+1|+C
Checking the result by deriving:
[x-1/ln(5)*ln|5^x+1|+C]’ = 1-1/ln(5)*1/(e^[x*ln(5)]+1)*e^[x*ln(5)]*ln(5)
= 1-5^x/(5^x+1) = (5^x+1-5^x)/(5^x+1) = 1/(5^x+1) everything okay!