Find the integral?

Question

Find the integral e^(2x)/(1+e^x) dx. 

 

Answer: e^x-ln(1+e^x)+C

 

 

Answer 1

I = ∫e^(2x)/(1+e^x) dx

I = ∫(e^x/(1+e^x)) e^x dx

u = 1+e^x; e^x = u-1

du = e^x dx

I = ∫((u-1)/u) du

I = ∫(1-1/u) du

I = u - ln|u| + K

u = 1+e^x > 0

I = 1+e^x - ln(1+e^x) + K

C = K + 1

I = e^x - ln(1+e^x) + C

check:

dI/dx = e^x - (1/(1+e^x))(e^x)

= e^x(1+e^x)/(1+e^x) - e^x/(1+e^x)

= (e^x + e^(2x) - e^x)/(1+e^x)

= e^(2x)/(1+e^x) √

Answer 2

This integral can be done via two u successive u substitutions. 

 

first let  y = e^x       then dx =  dy/y    so the integral becomes

 

Integral [ y² /(1 + y) ] (dy/y)  = Integral [y/(1+y)] dy

 

then let z = 1 + y        then dy = dz and the integral becomes

 

 Integral [ (z -1)/z] dz   =   Integral 1 dz - Integral [1/z] dz =   z - ln(z)  + CC

 

substitute back z =  1 + e^x    to get

 

  Integral =   1 + e^x - ln(1 + e^x ) +CC    , since CC is a constant we can let C = 1 +CC  and then

 

   Integral = e^x  -ln(1 + e^x)  +C

    

Answer 3

Sun,

          Hint: let u=e^x so du=e^xdx so the integrand is udu/(1+u) now divide u by 1+u and get 1-1/(1+u) so the integral becomes ∫(1-1/(1+u))du now integrate term by term and we get u-ln(1+u) then put u=e^x and add the constant.