Integration by parts VS. Formula in text

[Saladsamurai]

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

 

[Saladsamurai]

I am getting closer, but I must have screwed up a sign or something in the Int by parts...but I can't see where ...

 

[rocomath]

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

 

[Saladsamurai]

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...

 

[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 +.

 

[rocomath]

it checks out after taking the derivative.

 

[Saladsamurai]

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 +.

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

 

[Saladsamurai]

it checks out after taking the derivative.

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

Casey

 

[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 because i suck, but is all good.

 

[Saladsamurai]

My goal though is to get what I got in post #4 to look like the formula from the textbook 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 ?

 

[Saladsamurai]

why can't I factor this??!

 

[rocomath]

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

 

[Saladsamurai]

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

But it is part of the assignment!

 

[rocomath]

But it is part of the assignment!

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?

 

[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.

 

[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.

 

[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

should read

\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.