How can I integrate e^(3x)/(1+e^x)?

Question

Steps Please!

Answer 1

let u = e×.
Then du = e× dx

∫e^(3x)/(1 + e×) dx= ∫(e×)²/(1 + e×) * e× dx
= ∫u²/(1 + u) du

Using polynomial long division*
u²/(1 + u) = u - 1 + 1/(1 + u), so the integral becomes
∫u - 1 + 1/(1 + u) du
= u²/2 - u + ln(1 + u) + C

Substituting u = e× back in,
= (e^(2x))/2 - e× + ln(1 + e×) + C

We don't need to worry about negative values, since e× is always positive.

*We can also use the following manipulation:
u²/(1 + u) = (u² - 1 + 1)/(u + 1) =
(u² - 1)/(u + 1) + 1/(u + 1) =
(u + 1)(u - 1)/(u + 1) + 1/(u + 1) =
u - 1 + 1/(u + 1)

Answer 2

If you put e^x = u then e^x dx = du and dx =du/u and e^3x = u^3 Now you have Int u^3/(1+u) *du/u = Int u^2/(1+u)du =
Int[(u-1)+ 1/(u+1)] du = 1/2(u-1)^2+ lnI 1+uI +C
you can make the reverse substitution

Answer 3

Integrate e^(3x)/(1 + e^x).

First let's do a little algebra to make the expression easier to integrate.

e^(3x)/(1 + e^x) = e^(2x)[(1 + e^x) - 1]/(1 + e^x)
= e^(2x) - e^(2x)/(1 + e^x)
= e^(2x) - e^x[(1 + e^x) - 1]/(1 + e^x)
= e^(2x) - e^x + e^x/(1 + e^x)

Now we can integrate.

∫{e^(3x)/(1 + e^x)}dx

= ∫{e^(2x) - e^x + e^x/(1 + e^x)}dx

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

Answer 4

You can do this on the graphing calculator on TI-83.
Push Y1 button on the calculator and the calculator screen should say Y1=(Enter the above equator here) and push the graph button. This should give you a graph of it and then find the intersection of it to get the answer by using the trace function.