Question

Just how to start integral of ln(1+x^3)?
Medal~always

Answer 1

@waterineyes @johnweldon1993 @Jhannybean

Answer 2

Looks gross lol, I would do it by parts but I can see it'll probably get lengthy

u = ln(1 + x^3)

dv = dx

lets see if Jhanny has any better ideas :)

Answer 3

\[\int \ln(1+x^3)dx\] Yeah.. I would try it by parts too.

Answer 4

Wow I am trying to solve it and it's extremely long -.-

Answer 5

@johnweldon1993 is right, working on it by parts works the best.

Answer 6

Eww this is terrible lol...eventually looks like it might need a trig sub too

Answer 7

It says to approximate (from 0 to 0.5 for bounds)
but I'll just ask my prof tomorrow

Answer 8

thanks both

Answer 9

u = ln (1+x^3) dv = dx

du = \(\frac{\sf 3x^2dx}{\sf 1+x^3}\) v = x

Answer 10

\[\int_0^{1/2} \ln(1+x^3)dx = x\ln(1+x^3) -\int_0^{1/2}x\cdot \frac{3x^2}{1+x^3}dx\]

Answer 11

For the integral part you get : \[\int \frac{x^3}{x^3+1}\] But I don't remember how to simplify this... I recall you can split it up by long division or something like that, but I forgot how to do long division -.-

Answer 12

oop, I left out the 3 after factoring it out of the integral, typo.

Answer 13

Ohh, you might have to expand it in order to solve for it. haha. Duh?

Answer 14

oh crap nvm, i read it wrong and said the wrong thing

Answer 15

I'm gonna delete all of it cause it is incorrect info

Answer 16

so let's see if we let u=1+x^3 du=3x^2 yes?

Answer 17

so then we have \[\int (3x^2)^-1 lnu du\}

Answer 18

\[\int (3x^2)^{-1} ~ln~u ~~du\]

Answer 19

now it is a candidate for integration by parts

Answer 20

LIPETs. log > Inverse > polys> expos> trig

Answer 21

? what's that for?

Answer 22

u = ln(u) du = 1/u du

dv = (3x^2)\(^{-1}\) v = ...

Answer 23

LIPETs is what I use to figure out which function I place as u and which one i place as dv

Answer 24

oh, never seen it before, nice

Answer 25

But I think this case might me more complicated, so order will be switched.

Answer 26

hmm, I would use that

Answer 27

because integrating lnu, is not something I'd want to do

Answer 28

hmm... integrating dv.

Answer 29

it's messy

Answer 30

because I didn't put it in terms of u... nitwit

Answer 31

u=1+x^3 right, means x=(u-1)^1/3

Answer 32

should make it better

Answer 33

but @adamaero, why don't you give it a shot

Answer 34

no, way. Thanks all, but I've got other studying to do for everything else, and haven't started our current chapter yet, so on to that