Integration by parts VS. Formula in text

1. Oct 28, 2007

I am either making the same mistake repeatedly, or I can't factor!

I did this $$\int\ln(2x-3)dx$$ first by parts and then using the formula $$\int\ln u du=u\ln u-u+C$$

By parts I got:
$$u=\ln(2x-3)$$ dv=dx

so $$=x\ln(2x-3)-\int\frac{2x}{2x-3}dx$$ and by long division:

$$=x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx$$

$$=x\ln(2x-3)-x-\frac{3}{2}\ln(2x+3)+C$$

but I can't get that to match the formula result of

$$(2x-3)\ln(2x-3)-(2x+3)+C$$

Is the by parts correct?

Casey

2. Oct 28, 2007

I am getting closer, but I must have screwed up a sign or something in the Int by parts.....but I cant see where .......

3. Oct 28, 2007

rocomath

that's what i got, let me take the derivative and to check.

Last edited: Oct 28, 2007
4. Oct 28, 2007

Okay, so I have fixed a sign error and I get from parts:

$$x\ln(2x-3)-x-\frac{3}{2}\ln(2x-3)$$ and then multiplying thru

by 2 I get $$2x\ln(2x-3)-2x-3\ln(2x-3)$$ which is extremely close...but I can't figure out how to factor it properly....

Almost there :)

I just can't figure out to do with that -2x in the "middle" its a term not a factor......

5. Oct 28, 2007

neutrino

For one,

$$x\ln(2x-3)\int [1+\frac{3}{2x+3}]dx$$

should have a minus sign within the integral.

In the following step, you've multiplied the - outside the integral(that's supposed to be there ) with that +.

6. Oct 28, 2007

rocomath

it checks out after taking the derivative.

7. Oct 28, 2007

I don't follow. If I do that integral off to the side I get x+(3/2)ln(2x-3)

and then i distibuted the - sign.....

EDIT: I see, I just forgot to put it in the 1st post...scroll down to post#4

Last edited: Oct 28, 2007
8. Oct 28, 2007

What does? from which post#. I can't get it to match the formula.

Casey

9. Oct 28, 2007

rocomath

first post, except that the last ln should be ln(2x-3)

i hate the tables, i prefer hand-method :p takes too long and comes with errors cuz i suck, but is all good.

10. Oct 28, 2007

My goal though is to get what I got in post #4 to look like the formula from the text book which looks like this $$(2x-3)\ln(2x-3)-(2x+3)+C$$

but I have $$2x\ln(2x-3)-2x-3\ln(2x-3)+C$$

which is close...how does that factor?

If I factor that don't I just get $$(2x-3)\ln(2x-3)-2x+C$$ ?

Last edited: Oct 28, 2007
11. Oct 28, 2007

why can't I factor this:rofl:??!

12. Oct 28, 2007

rocomath

lol that's really hard ... quit obsessing!

Last edited: Oct 28, 2007
13. Oct 28, 2007

But it is part of the assignment!

14. Oct 28, 2007

rocomath

to put it in table form? eek!

i got:

$$\ln{(2x-3)}[2x-3]-x$$ ...

but idk what to do after that. and how did you get -2x?

15. Oct 28, 2007

neutrino

Maybe it's because what you're trying to find is an indefinite integral, and the answers to different methods employed usually differ by a constant.

16. Oct 29, 2007

coomast

This might help:

$$-x+C=-\frac{1}{2}(2x)+C=-\frac{1}{2}(2x-3)+C-\frac{3}{2}=-\frac{1}{2}(2x-3)+C_1$$

In which $$C_1=C-\frac{3}{2}$$ another constant is.

17. Oct 29, 2007

Curious3141

salad, I can't quote your post, browser issues. But your "book formula" integral is in error.

$$(2x-3)\ln(2x-3)-(2x-3)+C$$

$$\frac{1}{2}[(2x-3)\ln(2x-3)-(2x-3)]+C$$

(I corrected your sign error)

You cannot in general just use a formula for an integral of f(u) with a direct substitution of u=g(x) into the result. This is because what you are actually doing with a proper substitution is

$$\int{f(g(x))}dx = \int{\frac{dx}{du}[f(u)]}du = \int{\frac{f(u)}{u'}}du$$

In this case, u'(x) = 2, so you should divide the final expression by two.

I did not check your integration by parts, but the above correct answer for the formula method tallies with it, except for a constant term :

$$x\ln{(2x-3)} - x - \frac{3}{2}\ln{(2x-3)} = \frac{1}{2}(2x)\ln{(2x-3)} - \frac{1}{2}(2x) - \frac{1}{2}(3\ln{(2x-3)}) = \frac{1}{2}[(2x-3)\ln{(2x-3) - 2x}] = \frac{1}{2}[(2x-3)\ln{(2x-3) - (2x - 3)}] + C$$

which all works out.

Last edited: Oct 29, 2007