Question

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

Answer 1

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

Answer 2

yes

Answer 3

that is not integration by parts

Answer 4

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

Answer 5

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)- \(\int 2xdx\)= \(2x^2 +x -x^2 +C= x^2 +x+C\)

Answer 6

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

Answer 7

hehehe... sometimes, the prof tests students.

Answer 8

So sorry, I left out the ln

Answer 9

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

Answer 10

Losers example would still fit

but let u = ln(2x+1) and dv = dx

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

do I do integration by parts again?

Answer 15

yes, again

Answer 16

u=2x+1 or u=x?

Answer 17

whats a good rule of thumb when choosing u?

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

is

Answer 23

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

Answer 24

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

Answer 25

a two where?

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

:)

Answer 30

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

Answer 31

Thank you.

Answer 32

youre welcome

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

@amistre64

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

\[\int sec(x)~dx\]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
\[ln(sec(x)+tan(x))+C\]

Answer 43

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

Answer 44

sorry and ty

Answer 45

@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.

Answer 46

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

Answer 47

that e is not spose to be hanging out there lol

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

ok

Answer 52

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

Answer 53

its a form of it yes

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

:) good luck

Answer 58

I appreciate it.