Question

evaluate

Answer 1

\[\large \int\limits (x^{3m}+x^{2m} +x^m)(2x^{2m}+3x^m+6)^{1/m}dx,x>0\]

Answer 2

@experimentX

Answer 3

can you evaluate this without limits of integration ?

Answer 4

the question says for any natural number m evaluate the above integral

Answer 5

@satellite73

Answer 6

could you find a sum of thee integrals , each using the integration by parts?

Answer 7

nt too good in that part

Answer 8

\[=\int\limits x^{3m}(2x^{2m}+3x^m+6)^{1/m}\text dx+\int x^{2m}(2x^{2m}+3x^m+6)^{1/m}\text dx \]\[+\int x^m(2x^{2m}+3x^m+6)^{1/m}\text dx\]

Answer 9

now?

Answer 10

im am not sure if this is going to work but
\[\int uv^\prime=[uv]-\int vu^\prime\]

Answer 11

oh thats pretty long !!

Answer 12

any other idea?

Answer 13

i just have an intution that there is an easier way out

Answer 14

your probably right but i that is the best i can think of right now

Answer 15

\[u=x^m\]

Answer 16

from there you might be able to see what to do...it's not that bad

Answer 17

@apoorvk , @Callisto

Answer 18

@zarkon arent the coefficients and exponents showing some kind of symmetry?

Answer 19

@Zarkon and then we let
\[t=2u^3+3u^2+6u\]
?

Answer 20

yes

Answer 21

hmm lemme try that

Answer 22

but what do i do with 1/m exponent?

Answer 23

nothing special to do with it because finally u have
\[I=\frac{1}{6} \int\limits t^{1/m} dt \]

Answer 24

oh i see thx !!

Answer 25

are you missing an m

Answer 26

yes

Answer 27

1/(6m) ***