How do you intergrate e^3x/1+e^x?

Question

The question says you have to use the substitution u=1+e^x
I've substituted to get
e^3x/u^2-u but I don't know how to get rid of the e^3x???

Answer 1

I don't understand your steps. Transform it like this:



∫(eˣ)³dx/(1+eˣ) = ∫((eˣ)²)*eˣ*dx/(1+eˣ)



Now let 1+eˣ = u and you have eˣdx = du and eˣ = u-1, therefore



∫(u-1)²*du/u = ∫(u²-2u+1)*du/u = ∫udu - ∫2du + ∫du/u = u²/2 - 2u + ln|u| + c



Going back to x:



(1+eˣ)²/2 - 2(1+eˣ) + ln(1+eˣ) + c

Answer 2

int [ e^( 3x ) / ( 1 + e^x ) dx ]

Let u = 1 + e^x

du = e^x dx



int [ e^(2x ) e^x dx / ( 1 + e^x ) ]

= int [ (u - 1 )^2 du / u ]

= int [ ( u^2 - 2u + 1 ) du / u ]

= int [ ( u - 2 + 1 / u ) du ]

= ( 1 / 2 )u^2 - 2u + ln abs ( u ) + C

= ( 1 / 2 )( 1 + e^x )^2 - 2(1 + e^x ) + ln abs ( 1 + e^x ) + C

Answer 3

∫ e^3x/ (1+e^x) dx



Let u = 1+e^x

du = e^x dx

e^x = u-1

e^2x = (u-1)^2



∫ e^3x/ (1+e^x) dx = ∫ e^2x e^x / (1+e^x) dx = ∫ (u-1)^2 /u du

= ∫ (u^2-2u+1) /u du

= ∫ u du - 2 ∫ du + ∫ du/u

= u^2 /2 - 2u + ln(u) + C

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

Answer 4

Evaluate ∫[e^(3x)/(1+e^x)]dx



Let

u = (1+e^x)

(u - 1) = (e^x)



du = (e^x)dx

(1/(e^x))du = dx



thus



∫[e^(3x)/(1+e^x)]dx = ∫[e^(3x)/u](1/(e^x))du



∫[e^(3x)/(1+e^x)]dx = ∫[e^(3x-x)/u]du



∫[e^(3x)/(1+e^x)]dx = ∫[e^(2x)/u]du



∫[e^(3x)/(1+e^x)]dx = ∫[(e^x)²/u]du



recall (u - 1) = (e^x)



∫[e^(3x)/(1+e^x)]dx = ∫[(u - 1)²/u]du



expand: (u - 1)² = (u² - 2u + 1)



∫[e^(3x)/(1+e^x)]dx = ∫[(u² - 2u + 1)/u]du



∫[e^(3x)/(1+e^x)]dx = ∫[u²/u - 2u/u + 1/u]du



∫[e^(3x)/(1+e^x)]dx = ∫[u - 2 + 1/u]du



∫[e^(3x)/(1+e^x)]dx = ∫[u]du - ∫[2]du + ∫[1/u]du



∫[e^(3x)/(1+e^x)]dx = (1/2)u² - 2u + (ln u) + C



recall u = (1+e^x)



∫[e^(3x)/(1+e^x)]dx = (1/2)(1+e^x)² - 2(1+e^x) + (ln (1+e^x)) + C <--- answer



Note simplify further if you want to.