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1) If you really want to use parts (as opposed to simply a u-sub with u=3x+1), the product (after bringing out the 5) is \(\displaystyle \mbox{ \frac{1}{\sqrt{3x + 1}} \times 1}\).

2) Nicely done.

3) The product is \(\displaystyle \mbox{ \ln{(2x)} \times 1}\).

4) Nice.

5) Good.

6) Does \(\displaystyle \mbox{\int \frac{1}{x} dx - \int \frac{1}{x} dx = 0 ?\)
 
1) Wonderful.

3) You lost the 2 but good otherwise.

6) Look very closely at your (correct) response to his hint.
 
3) Keep the 2 where it was. u = ln(2x), dv/dx = 1.

6) So, \(\displaystyle \mbox{ \int \frac{1}{x} dx - \int \frac{1}{x} dx = ?}\)
 
3) You're beginning to worry me. ln(2x) does not equal ln(x) * ln(2).

All I am saying: let u=ln(2x) and dv/dx = 1.

It is virtually the same as what you did with u=ln(x) and dv/dx = 1.

6) Although you are technically correct, the point is that the difference is some constant.

1 is just a constant so is enveloped by the constant of integration.
 
Sorry but I was wondering what you did in the last two steps of problem five. I was getting confused as to what happened to the final " S cos^2 x dx " and where the " 1/2 " came from in the final line. Perhaps this is a really obvious step but I am having trouble seeing it tonight.
 
Bit of a trick, Legacy: add \(\displaystyle \mbox{\int \cos^2{(x)} dx}\) to both sides of the second-to-last line.
 
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