Question

how to solve this integral question? integral of
[(x power 7m+x power 2m+x power m)].[(2x power 6m+7x power m+14)] power 1/m

Answer 1

Is this right?
\[\int\limits_{}^{}(x^{7m} + x^{2m} + x^m)*(2x^{6m} + 7x^m + 14)^{1/m}\]

Answer 2

ya

Answer 3

Is the whole equation to the power 1/m, or just the second term?

Answer 4

only the second term...

Answer 5

This isn't going to be easy by hand, if at all possible.

Answer 6

i suppose we are integrating with respect to x right?

Answer 7

ya i knw dat...i did dat but amnt getting it..

Answer 8

Yeah, I don't think there is an easy way to solve this by hand. Where did the problem come from?

Answer 9

this problem is above me lol

Answer 10

wait
did you try integration by parts?

Answer 11

maybe we can find a pattern and right it as a sum

Answer 12

write*

Answer 13

Integration by parts is useless; Since we don't know what m is, there's no way to reduce the degree of the polynomial.

Answer 14

we can see if there is a pattern though and write it as a sum

Answer 15

yea i dont think it will work though

Answer 16

its from tata mc grawls...its realy tough

Answer 17

all the problem lies in m ...
we jus assume that m is one ...
that way i can tell what the answer is but with m its too complicated

Answer 18

@myininaya "we can see if there is a pattern though and write it as a sum"

Some fiddling about with Wolfram does appear to give a pattern for the integral:

mx^(m-1)[Ax^2m + Bx^6m + Cx^7m + D x^12m + Ex^m + F) all over

m(2x^6m + 7x^m + 14) ^ (m-1/m) where A to F are given by

A = 35 + 14(m-2)
B = 212 +100(m-2)
C = 125 +53(m-2)
D = 40 14(m-2)
E = 77 +35(m-2)
F = 28+14(m-2)

So you could substitute these in to get the thing but I am feeling too lazy to do that.

My guess is Wolfram is using some kind of series calculations so I don't really know how accurate A to F are, maybe more time with the computation would produce something a bit different.

Answer 19

yeah i was pretty lazy too about this problem mostly because it looks too ugly

Answer 20

guys please explain me this question please??

Answer 21

I thought we did?

Answer 22

I just put an answer above...

Answer 23

estudier please explain me step by step please...?

Answer 24

i recall some place to use a "z sub" where z is an exponential term that plays nicely with others ..

Answer 25

wats dat?

Answer 26

cant recall the details a tthe moment. but they take a common factor of the exponents and use a term that can make the work easier by substitution. whether it works here or not i dunno

Answer 27

wat is z sub?

Answer 28

its the same as "h sub" but with a z :)

like i said, i cant recall the details that well

Answer 29

oh ok

Answer 30

I used Wolfram to integrate the expression starting with 1/2 then 1/3, 1/4 etc.

From this you can see a pattern in the results from which you can obtain a general result inn m.

Answer 31

Perhaps there is a tricky way to do it analytically but I don't know it.

Answer 32

(x^7m^2 +x^2m^2 +1^m) (2x^6m +7x^m +14)^1/m

x^14m +x^42m +1^m
2x^6m +7x^m +14
--------------------
2x^(20m) +2x^(50m) +2x^(7m)
7x^(15m) +7x^(43m) +7x^(2m)
+14x^(14m) +14x^(42m) +14x^(m)


--------------------------------------------
| 2x^(20m) +2x^(50m) +2x^(7m) | ^1/m
| 7x^(15m) +7x^(43m) +7x^(2m) |
| +14x^(14m) +14x^(42m) +14x^(m) |
----------------------------------------------

I got know idea what im doing at the moment :)

Answer 33

u know what happened to Hamilton and the Quaternions, Cantor and his infinities....

Answer 34

me? aint got clue

Answer 35

Joke...you will go a bit mad..:-)

Answer 36

That's not USA mad that's Brit mad.

Answer 37

:)

Answer 38

whta u can do is ...take x common in the first part of the question....then take that x into the part which has power .... then substitute the part under power ... u will see the other part gets canceled .....then a simple integration would work

Answer 39

You lost me....

Answer 40

u lost me???

Answer 41

You might be right, I'd like to see it.

Answer 42

i believe he said factor out an x^m and up it to the mth power and insert it into the mthroot

Answer 43

i kinda did that by a mth-ed up the whole thing

Answer 44

no put x into 1/m ..then it becomes x^m

Answer 45

thats what I said :)

Answer 46

sry i understood it wrong

Answer 47

OK, so what's the integral then?

Answer 48

\[\int\limits_{}^{}(x^{7m-1} + x^{2m-1} + x^{m-1})*(2x^{7m} + 7x^{2m} + 14x^{m})^{1/m}\]

Answer 49

1/14m{ [(2x^7m + 7x^2m +14x^m)^1/m +1 ] / 1/m + 1}

Answer 50

\[u=2x^{7m}+7x^{2m}+14x^{m}\]
\[1/14\int\limits_{?}^{?}u^{1/m}du\]

Answer 51

thats what i was trying to say... thank thomas

Answer 52

You deserve all the credit Ishaan :)

Answer 53

If you let x = 1/m, what cancels out?

Answer 54

The way I did it by hand was to first exponentiate everything in the second parentheses by 1/m. Then I expanded everything out. Then break up the integral and do separate u subs.

Answer 55

you cant let x =i/m ...x is a variable ........while m is const

Answer 56

its not i its 1

Answer 57

Oh, i read up a few lines you said to do that?

Answer 58

no i said to take x common out of the first expression........and then put it under power 1/m

Answer 59

OH.