Mathematics OpenStudy (anonymous):

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

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

yes OpenStudy (amistre64):

that is not integration by parts OpenStudy (amistre64):

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 OpenStudy (loser66):

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)- $$\int 2xdx$$= $$2x^2 +x -x^2 +C= x^2 +x+C$$ OpenStudy (amistre64):

.... yeah, i spose thats technically valid ;) OpenStudy (loser66):

hehehe... sometimes, the prof tests students. OpenStudy (anonymous):

So sorry, I left out the ln OpenStudy (anonymous):

$\int\limits_{}^{}\ln(2x+1)dx$ OpenStudy (amistre64):

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

I tried that already and didn't come up with the correct answer OpenStudy (amistre64):

walk me thru what you tried ... the error usually tends to be a simple algebra issue OpenStudy (anonymous):

I think Im stuck at $xln(2x+1)-\int\limits_{}^{}x*\frac{ 1 }{ (2x+1) }dx$ OpenStudy (anonymous):

do I do integration by parts again? OpenStudy (amistre64):

yes, again OpenStudy (anonymous):

u=2x+1 or u=x? OpenStudy (anonymous):

whats a good rule of thumb when choosing u? OpenStudy (amistre64):

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

how do I know to use the 1 over the poly and not just the poly? OpenStudy (amistre64):

polynomials are good u choices; then lns, es, trig stuff. I used to know an anacronym li ILATE or something OpenStudy (amistre64):

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

is OpenStudy (amistre64):

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

is$\int\limits_{}^{}\ln(2x+1)dx =\frac{ 1 }{ 2x+1 }?$ OpenStudy (anonymous):

a two where? OpenStudy (amistre64):

$xln(2x+1)-\int\limits_{}^{}\frac{ 2x }{ 2x+1 }dx$ $xln(2x+1)-\int\limits_{}^{}\frac{ 2x+1-1 }{ 2x+1 }dx$ $xln(2x+1)-\int\limits_{}^{}\frac{ 2x+1}{2x+1}-\frac{1 }{ 2x+1 }dx$ $xln(2x+1)-\int\limits_{}^{}1-\frac{1 }{ 2x+1 }dx$ might help to clear up some clutter OpenStudy (amistre64):

$ln(2x+1)\to\frac{(2x+1)'}{2x+1}=\frac{2}{2x+1}$ OpenStudy (anonymous):

oh right right, now that I see the +1 -1 thing I just remembered my teacher did this one in class OpenStudy (amistre64):

:) OpenStudy (anonymous):

it would have never occurred to me to do the +1 -1 OpenStudy (anonymous):

Thank you. OpenStudy (amistre64):

youre welcome OpenStudy (amistre64):

adding zero and multiplying by 1 are a mathemticians best tricks ;) OpenStudy (anonymous):

I know, I just never recognize the right time to use it. OpenStudy (anonymous):

amistre64 can you do the problem for me because I just did it and I get the wrong answer OpenStudy (anonymous):

@amistre64 OpenStudy (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 OpenStudy (amistre64):

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

OMG!!! I forgot the + 0 within 2 seconds. I need a break OpenStudy (amistre64):

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

Trust me I completely understand. It seems I forget everything I previously learned with every new lesson OpenStudy (amistre64):

$\int sec(x)~dx$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 $ln(sec(x)+tan(x))+C$ OpenStudy (amistre64):

*multiply by a useful form of 1 .... my fingers hate me :/ OpenStudy (anonymous):

sorry and ty OpenStudy (anonymous):

@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. OpenStudy (amistre64):

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 OpenStudy (amistre64):

that e is not spose to be hanging out there lol OpenStudy (anonymous):

Okay, I understand. I just dont see how I will burn the rule into my brain. OpenStudy (amistre64):

if we use the textbook method of just substitution; it takes longer but gets to the same point OpenStudy (amistre64):

$\int \frac{1}{2x+1}dx$ $u=2x+1~;~du=2~dx$$\frac{1}{2}du=dx$sub the parts $\int \frac12\frac{1}{u}du$ OpenStudy (anonymous):

ok OpenStudy (anonymous):

the book says that the answer is $\frac{ 1 }{ 2 }(2x+1)\ln(2x+1)-x+c$ is this the same as your answer? OpenStudy (amistre64):

its a form of it yes OpenStudy (anonymous):

I don't really see it. I believe you of course. I just dont see it OpenStudy (amistre64):

$xln(2x+1)-x+\frac12ln(2x+1)+C$ $xln(2x+1)+\frac12ln(2x+1)-x+C$ factor out the ln(2x+1) $(x+\frac12)ln(2x+1)-x+C$and factor out a 1/2 $\frac12(2x+1)~ln(2x+1)-x+C$ OpenStudy (anonymous):

Oh, thank you for going out your way to show that to me. Im finally all done. thanks a bunch. 1000 points :) OpenStudy (amistre64):

:) good luck OpenStudy (anonymous):

I appreciate it.

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