Integration by part Please integrate (2x+1)dx

Hi there dizzie! :) This is what we have to solve?\[\Large \int\limits(2x+1)\;dx\]

yes

that is not integration by parts

integration by parts is useful when defining a product; what you have is more along the lines of: the integration of a sum, is the sum of integrations

if you still want to take integral by part, it's ok, too. let u = (2x+1) du = 2 dx, and dv = dx , v =x apply formula you have x(2x+1)- \(\int 2xdx\)= \(2x^2 +x -x^2 +C= x^2 +x+C\)

.... yeah, i spose thats technically valid ;)

hehehe... sometimes, the prof tests students.

So sorry, I left out the ln

\[\int\limits_{}^{}\ln(2x+1)dx\]

Losers example would still fit but let u = ln(2x+1) and dv = dx

I tried that already and didn't come up with the correct answer

walk me thru what you tried ... the error usually tends to be a simple algebra issue

I think Im stuck at \[xln(2x+1)-\int\limits_{}^{}x*\frac{ 1 }{ (2x+1) }dx\]

do I do integration by parts again?

yes, again

u=2x+1 or u=x?

whats a good rule of thumb when choosing u?

u = 1/(2x+1), otherwise your just rocking back and forth

how do I know to use the 1 over the poly and not just the poly?

polynomials are good u choices; then lns, es, trig stuff. I used to know an anacronym li ILATE or something

because the function is 1/(2x+1) ; otherwise youd just be doing a substitution run and not a by parts run

is

and you have a slight error in the latex, should be a 2 up top\[xln(2x+1)-\int\limits_{}^{}x*\frac{ 2 }{ (2x+1) }dx\]

is\[\int\limits_{}^{}\ln(2x+1)dx =\frac{ 1 }{ 2x+1 }?\]

a two where?

\[xln(2x+1)-\int\limits_{}^{}\frac{ 2x }{ 2x+1 }dx\] \[xln(2x+1)-\int\limits_{}^{}\frac{ 2x+1-1 }{ 2x+1 }dx\] \[xln(2x+1)-\int\limits_{}^{}\frac{ 2x+1}{2x+1}-\frac{1 }{ 2x+1 }dx\] \[xln(2x+1)-\int\limits_{}^{}1-\frac{1 }{ 2x+1 }dx\] might help to clear up some clutter

\[ln(2x+1)\to\frac{(2x+1)'}{2x+1}=\frac{2}{2x+1}\]

oh right right, now that I see the +1 -1 thing I just remembered my teacher did this one in class

:)

it would have never occurred to me to do the +1 -1

Thank you.

youre welcome

adding zero and multiplying by 1 are a mathemticians best tricks ;)

I know, I just never recognize the right time to use it.

amistre64 can you do the problem for me because I just did it and I get the wrong answer

@amistre64

int ln(2x+1) x ln(2x+1) - int 2x/(2x+1) dx x ln(2x+1) - int (2x+1-1)/(2x+1) dx x ln(2x+1) - int 1 -1/(2x+1) dx x ln(2x+1) - int dx - int 1/(2x+1) dx x ln(2x+1) - x - 1/2 int 2/(2x+1) dx x ln(2x+1) - x - 1/2 ln(2x+1) + C (x- 1/2) ln(2x+1) - x + C

opps, i ran over a negative \[xln(2x+1)-\int\limits_{}^{}1-\frac{1 }{ 2x+1 }dx\] \[xln(2x+1)-\int dx\color{red}{+}\frac12\int \frac{2}{ 2x+1 }dx\] \[xln(2x+1)-x+\frac12ln(2x+1)+C\]

OMG!!! I forgot the + 0 within 2 seconds. I need a break

when I saw the integration of sec(x) the first time i got sooo angry lol

Trust me I completely understand. It seems I forget everything I previously learned with every new lesson

\[\int sec(x)~dx\]multiply by a useful form of a \[\int sec(x)\frac{sec(x)+tan(x)}{sec(x)+tan(x)}~dx\] \[\int \frac{sec^2(x)+sec(x)tan(x)}{tan(x)+sec(x)}~dx\] now the top is the derivative of the bottom to ln up \[ln(sec(x)+tan(x))+C\]

*multiply by a useful form of 1 .... my fingers hate me :/

sorry and ty

@amistre64 in step 5 how did you get 1/2 and int 2/(2x+1)? sorry :l I just hate to have all your typing go to waste because I can get past this step.

lets split the setup again,\[\frac{2x+1-1}{2x+1}\] keep in mind how we got e/(2x+1) to start with; we took the derivative of ln(2x+1) \[\frac{2x+1}{2x+1}-\frac{1}{2x+1}\] \[1-\frac{1}{2x+1}\] since we already know that 2/(2x+1) integrates back into ln(2x+1), lets multiply that by 2/2 (a useful from of 1) \[1-\frac22\frac{1}{2x+1}\] \[1-\frac12\frac{2}{2x+1}\] now were set

that e is not spose to be hanging out there lol

Okay, I understand. I just dont see how I will burn the rule into my brain.

if we use the textbook method of just substitution; it takes longer but gets to the same point

\[\int \frac{1}{2x+1}dx\] \[u=2x+1~;~du=2~dx\]\[\frac{1}{2}du=dx\]sub the parts \[\int \frac12\frac{1}{u}du\]

ok

the book says that the answer is \[\frac{ 1 }{ 2 }(2x+1)\ln(2x+1)-x+c\] is this the same as your answer?

its a form of it yes

I don't really see it. I believe you of course. I just dont see it

\[xln(2x+1)-x+\frac12ln(2x+1)+C\] \[xln(2x+1)+\frac12ln(2x+1)-x+C\] factor out the ln(2x+1) \[(x+\frac12)ln(2x+1)-x+C\]and factor out a 1/2 \[\frac12(2x+1)~ln(2x+1)-x+C\]

Oh, thank you for going out your way to show that to me. Im finally all done. thanks a bunch. 1000 points :)

:) good luck

I appreciate it.

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