Integrating: Calculus II Question?

Question

How would you BEGIN to integrate this problem? I do not need you to solve the problem for me. Thanks!

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

Answer 1

Let u=e^(-x)+1

After that it should be easy.

Edit: looking further into this problem, I realized it might not be easy, so I'll give one more hint: when you get the expression for dx in terms of u (think about how you can write e^x in terms of u) and make your substitution, you will end up with a (factored) rational function. At this point you should apply partial fractions to split it into two fractions. THEN it's easy.

Edit 2: After doing some more thinking about the problem, I realized there's a way to avoid the partial fractions: multiply both the numerator and denominator by e^x, then let u=1+e^x. You should then be able to simply cancel the numerator after the subsitution.

Answer 2

Integral (2 / (e^(-x) + 1) ) dx

Let u = e^(-x) + 1. Then
u - 1 = e^(-x)

Changing this to logarithmic form,

ln(u - 1) = -x, so
-ln(u - 1) = x

Differentiating both sides,

1/(u - 1) du = dx, so our integral becomes

Integral [ (2 / u) (1/(u - 1)) du ]

Pulling the constant 2 out, we have

2 * Integral [ 1 / (u(u - 1)) ] du

To solve this, we use partial fractions.

1 / [u(u - 1)] = A/u + B/(u - 1)

Multiplying both sides by u(u - 1), we get

1 = A(u - 1) + B(u)

Remember that this is true for ALL values of u.
Let u = 1: Then 1 = A(0) + B(1), so B = 1.
Let u = 0: Then 1 = A(0 - 1) + B(0), so A = -1.

Therefore,

2 * Integral [ 1 / (u(u - 1)) ] du

Is the same as

2 * Integral [ (-1/u) + 1/(u - 1) ] du

Which we can now integrate easily.

2 * [-ln|u| + ln|u - 1|] + C

2 * [ln|u - 1| - ln|u|] + C

But u = e^(-x) + 1, so we have

2 * [ln|e^(-x) + 1 - 1| - ln|e^(-x) + 1|] + C
2 * [ln|e^(-x)| - ln|e^(-x) + 1|] + C

Since ln and e are inverses of each other, ln|e^(-x)| = -x, so

2 * [-x - ln|e^(-x) + 1|] + C

Answer 3

If youy positioned lnx = u x=e^u and a million/x dx = du so dx = e^u du so that you get the crucial u^2 *e^u du Integrating through factors = u^2 e^u -2 Int u*e^u du and Int u*e^u du= = u*e^u -Int e^u du = e^u(u-a million) So the outcome is e^u( u^2 -2u +2) = x( (lnx)^2 -2 ln x +2) the same with the second one integrating first e^-5x it is -a million/5 e^-5x

Answer 4

Integrate 2/(e^(-x) + 1) with respect to x.

First, rearrange the terms to make them easier to integrate.

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

∫{2/(e^(-x) + 1)}dx = ∫{2e^x/(e^x + 1)}dx

Let u = e^x + 1
du = e^xdx
2du = 2e^xdx

∫{2e^x/(e^x + 1)}dx = ∫(2/u)du = 2ln(u) + C = 2ln(e^x + 1) + C

No absolute value signs are needed since e^x + 1 is always positive.

Answer 5

Have you tried a U substitution? Try letting U = e^(-x)+1, for example...

Answer 6

Use U substitution. :) Generally in U-substitution, you use the more complicated part of the problem, in this case, "e".

Answer 7

right u=1+e^(-x ) then chain rule