Question

Integration by parts.

x^4 ln(x) dx

u = ln(x)
du = 1/x

dv = x^4
v = (x^5)/5

Answer 1

\[\frac{x^5}{5} * \ln(x) - \int\limits \frac{x^5}{5} * \frac{1}{x}\]

Answer 2

looking good so far!

Answer 3

\[\frac{x^5}{5} * \ln(x) - \int\limits \frac{x^4}{5}\]

Answer 4

now integrate one more time

Answer 5

\[\frac{x^5}{5} * \ln(x) - \frac{x^5}{20} + c\]

Answer 6

20?

Answer 7

I could get rid of the first x^5 with the last right? 25 I meant.

Answer 8

Yes x^5/25

Answer 9

better haha

Answer 10

you can combine as one fraction

Answer 11

now could I do \[\ln(x) * \frac{5x^5}{25} - \frac{x^5}{25}\]

Answer 12

dont forget the constant lol..

Answer 13

\[\ln(x) * \frac{4x^5}{25}\] + c ??

Answer 14

err tyhat messed up.

Answer 15

hmm..it doesnt sound right...you can factor out x^5 though

Answer 16

remember you have ln x 5 x^5 - x^5 <---you cant combine because of ln x...it's like a variable

Answer 17

\[\large\frac{5x^5 (\ln x) - x^5 + c}{25}\]
see that?

Answer 18

oh yeah I forgot I have to put the /25 under both huh...

Answer 19

no...my point is 5x^5 (ln x) lol

Answer 20

Well is it the entire thing - x^5/25?

Answer 21

what do you mean?

Answer 22

\[(\ln(x) * \frac{x^5}{5}) -\frac{x^5}{25} +c?\]

Answer 23

didnt you make it 5x^5/25 to make it common denominator?

Answer 24

That's what I am being confused about.

Answer 25

what?

Answer 26

if I made it a common denom of 25, then it would be ln(x)/5

Answer 27

I wasn't sure if the entire right side (u*v) had parenthesis or if I could deal with each part individually.

Answer 28

hmmm let me put it this way \[\large \frac{ab}{5} = \frac{5ab}{25}\]
does that help?

Answer 29

Yes I understand that.

Answer 30

In this case there isnt' an '=' though.

Answer 31

But I guess what you were saying was.

Answer 32

\[\frac{\ln(x)x^5}{5} -\frac{x^5}{25} +c? would be ) * \frac{5(\ln(x)x^5)}{25} -\frac{x^5}{25} +c?\]

Answer 33

yes!!!

Answer 34

okies... :)

Answer 35

But before you wrote it as 5x^5 * ln(x) wouldn't it be destributed over both ln(x) and x^5? so 5ln(x) * 5x^5?

Answer 36

\[\large ab \implies ba\]
commutative property of multiplication
\[\large (\ln x) x^5 \implies x^5 (\ln x)\]

Answer 37

\[\LARGE 5(\ln x) x^5 \implies 5x^5 (\ln x)\]

Answer 38

5ab = 5*a*b ..... NOT 5*a * 5*b

Answer 39

it seems you are tired @KonradZuse ;) you're getting properties mixed up

Answer 40

with all these discussion of properties...we forgot the integration question..do you know it now?