Integral Help, a number of questions.?

Question

Hi, you guys are great help.

I'm reviewing for an exam I have in a few hours (it's costing me sleep, but I guess it's worth it), and I'm using a review "packet" which apparently has the same "format" of questions as on the actual test. (So in other words, there will be different values, but same set-up) Anyways, I've completed a few problems and I'm hoping if some of you can check 2, and show me how to do some others. Thanks.
I'm sorry if it seems like I'm overkilling it by including more than one question, but apparently I'm running out of questions I can ask.


Problems I've done, and need checked:
1) Integral of e^(1/x+1)/(x+1)^2 dx

For that, I got -e^(1/x+1)+c

2) Integral of 1/(x^2)(e^(2/x))dx
For that, I got -1/2(e^-2/x) + c

Questions I need full assistance with:

3) Evaluate Integral of e^x/(e^x + 1)^(1/2) dx

4.) Evaluate Definite Integral 1/x-1 dx; b(top):e+1 a(bottom):2

Thanks.

Answer 1

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

We're going to use substitution. I'll first rearrange the terms to show you how substitution applies.

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

Let u = 1/(x + 1). Then
du = -1/(x + 1)^2 dx Multiplying both sides by (-1), we get
(-1) du = 1/(x + 1)^2 dx. {Note: Look at how this is the tail end of our integral. This means that the tail end of our integral following the substitution will be (-1) du, and we get

Integral (e^(u) (-1) du ). Pulling the constant (-1) out of the inside, we get

(-1) Integral (e^u du)

Which we can now integrate easily.

(-1) e^u + C

Substituting u = 1/(x + 1) back, we get

(-1) e^[1/(x + 1)] + C

I'm going to jump to #3 and #4, since you need full assistance.

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

First, I'm going to rearrange the integral.

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

Now I'm going to use substitution.
Let u = e^x + 1. Then
du = e^x dx {Note: this is the tail end of our integral}.

So our substitution results in

Integral ( 1/u^(1/2) du)
Integral ( u^(-1/2) du)
Which is now an easy integral to solve; using the reverse power rule, we get

2u^(1/2) + C. But since u = e^x + 1, we get

2[e^x + 1]^(1/2) + C

4) Integral (2 to e+1, 1/(x - 1) ) dx

The integral of 1/(x - 1) is just ln|x - 1|. This is a relatively simple substitution which is not worth doing. Therefore, we get

ln|x - 1| {evaluated from 2 to e+1}.
If F(x) = ln|x - 1|, we want to evaluate F(e+1) - F(2).

F(e + 1) - F(2) = ln|e + 1 - 1| - ln|2 - 1|
= ln|e| - ln|1|

Note that ln(e) = 1, and ln(1) = 0, so our answer is
= 1 - 0
= 1