Question

What is the indefinite integration of:\[\int x4^x(3+ln(x^6)+3xln(4)ln(x))dx\]

Answer 1

i think this proplem is based on integration by parts

Answer 2

i spose it could be based on that. But this is one of my own creations :)

Answer 3

it even broke the wolf

Answer 4

@FoolForMath hows this for a problem of the day?

Answer 5

soft

Answer 6

Looks scary.

Answer 7

that it does, latex makes everything looks scary tho

Answer 8

I am going to scan my solution this is fun lol

Answer 9

how do you integrate x*lnx*4^x lol @amistre64

Answer 10

:) hint: expand the integrand out and simplify it all before trying to integrate

Answer 11

yeah i did that i got 3 parts i am doing the second part never encountered that integral before lol

Answer 12

i think i got it nvm lol

Answer 13

the 2nd one becomes

x 4^x ln(x^6)

6x 4^x ln(x)

Answer 14

yupp i got that just integrating my life away now lol

Answer 15

dont integrate them separately; look at it as a whole and try to think of a simpler idea that unites them

Answer 16

the ln(4) should be a dead give away to my evilness lol

Answer 17

yeah i saw that lol :P

Answer 18

almost finished do i get a hint in how to integrate lnx*4^x

Answer 19

@amistre64

Answer 20

i take it that is from the last term?

Answer 21

yeahh

Answer 22

\[x4^x3xln(4)ln(x)\]

is rather messy; but lets organize it better

\[3x^2\ ln(x)\ 4^xln(4)\]
\[[3x^2]\ [ln(x)]\ [4^xln(4)]\]

Answer 23

yeah so i did this by integration by parts and made the first 2 terms u and the last term dv

and then my du=2xlnx +x and dv=4^x

Answer 24

if we look at the exapnsion; notice the each term has similar parts\[3x4^x+6x4^xln(x)+3x^2ln(x)4^xln(4)\]
and recall the product rule:\[\frac{d}{dx}fg=f'g+fg'\]
this expands to more than 2 factors right?

im not sure what the by parts method would bring us, but its messy

Answer 25

everything is done i just neeed to know how to integrate lnx*4^x

Answer 26

do you know how to integrate lnx*4^x because i cant think of any way to integrate this in my past knowloedge lol :P if not i am just going to post my solution

Answer 27

i cant think of a way to integrate ln(x) 4^x unless we try parts

ln(x) 4^x ln(4) - {S} 4^x ln(4)/x dx

that doent seem to play nice does it

Answer 28

i did that backewards; i tend to derive the integrating part :/

Answer 29

nope then i get (4^x)/x how do we integrate that ?

Answer 30

by parts isnt a gaurentee that you get a nicer or easier function to play with is it?

Answer 31

int by parts is a technique, not a rule

Answer 32

yeah its a technique

Answer 33

as such, we mmay end up with a harder integration to play with and have to back track or go another route

Answer 34

the other route to me is to see the problem in a different manner

Answer 35

assume that the results follow from something that is easier known

3 terms with like parts COULD be a result of:\[\frac{d}{dx}fgh=f'gh+fg'h+fgh'\]
and then look for a way that this can match

Answer 36

we can even format that to a similar by parts formula

\[fgh=\int f'gh+\int fg'h+\int fgh'\]
\[\int fgh'=fgh-\int f'gh-\int fg'h\]

Answer 37

alright time to post it lol only one integral couldnt be solved lol other than that its okay @amistre64

Answer 38

that was a good effort at least :)

i devised this from the concept that:

\[\frac{d}{dx}3x^2\ 4^x\ ln(x)=3x'^2\ 4^x\ ln(x)+3x^2\ 4'^x\ ln(x)+3x^2\ 4^x\ ln'(x)\]
\[\frac{d}{dx}3x^2\ 4^x\ ln(x)=6x\ 4^x\ ln(x)+3x^2\ 4^xln(4)\ ln(x)+3x\ 4^x\]
\[\frac{d}{dx}3x^2\ 4^x\ ln(x)=x4^x(3+ln(x^6)+3x\ ln(4)\ ln(x))\]

Answer 39

@Aka_966