Question

Integration question..any idea? its below

Answer 1

\[\int\limits_{?}^{?} \frac{ x ^{5m-1 } +2x ^{4m-1}dx}{ (x ^{2m}+x ^{m}+ 1)^{3} }\]

Answer 2

indefinite integral

Answer 3

.

Answer 4

@IrishBoy123 ?? i didn't get u..?

Answer 5

oh ok

Answer 6

@ganeshie8 first of all we can simplifie this fraction - not ? and this ,,dx" from numerator not has place next the fraction line ?

Answer 7

- sorry - but i worked a long time ago with integration - but my first opinion is these above wrote

Answer 8

I'm thinking of partial fractions and also trying to simplify the denominator by writing it as \(\dfrac{1-x^{3m}}{1-x^m}\)

but these all look painful

Answer 9

its ok..thanks anyway

Answer 10

Still ugly, slightly pretty though:

\[\int \frac{x^{m-1} (x+2)}{(x^m+1+x^{-m})^3 }dx\]

Answer 11

That actually leads to a possibly nice substitution that may or may not lead anywhere, \(x^m=u\)

\[\frac{1}{m} \int \frac{ u^{1/m}+2}{(u+1+u^{-1})^3 }du\]

Answer 12

Divide \(x^{6m}\) top and botttom

Answer 13

@Empty i think u have made a mistake in the numerator.. its (x^m+2)..right?

Answer 14

\[\int \frac{ x ^{5m-1 } +2x ^{4m-1}dx}{ (x ^{2m}+x ^{m}+ 1)^{3} } =\int \frac{ x ^{-m-1 } +2x ^{-2m-1}dx}{ (1+x ^{-m}+ x^{-2m})^{3} } \]

Answer 15

see the obvious substitution ?

Answer 16

cool!!!

Answer 17

if we first multiply by x/x you get
\[ \frac{x^{5m}+2x^{4m} }{x(x^{2m}+x^m+1)^3} \ dx\]
now use Empty's sub:
\[ \frac{ u^5 + 2u^4}{x(u^2+u+1)^3} \frac{1}{m} \frac{1}{x^{m-1}} \ du \\
\frac{1}{m}\frac{1}{u}\frac{ u^5 + 2u^4}{(u^2+u+1)^3} \ du
\]
and then
\[ \frac{1}{m} \int \frac{u^3(u+2)}{(u^2+u+1)^3 } \ du \]

Answer 18

ya !! i think i got a clue ..i will try and tell u!!

Answer 19

I think we need to be careful when dividing top and bottom, we may not always get the correct answer

Answer 20

Here were my steps:
\[\int \frac{ x ^{5m-1 } +2x ^{4m-1}dx}{ (x ^{2m}+x ^{m}+ 1)^{3} } = \int \frac{x^{4m-1}( x ^{m } +2) dx}{ (x^m)^3(x ^{m}+1+x ^{-m})^{3} }\]
\[=\int \frac{x^{m-1} (x+2)}{(x^m+1+x^{-m})^3 }dx\]

Answer 21

But yeah, this is old news just sorta showing @priyar who was asking

Answer 22

i got:
\[1/2m (\frac{ 1 }{( x ^{-2m}+x ^{-m}+1)^{2} })\]

Answer 23

is this correct? @ganeshie8 ?

Answer 24

You should never ask that question. Differentiate and see if you get back the integrand :)

Answer 25

http://www.wolframalpha.com/input/?i=differentiate+1%2F%282m%29*x%5E%284m%29+%2F+%281%2Bx%5Em%2Bx%5E%282m%29%29%5E2

Looks we're good ! :)

Answer 26

\[1/2m (\frac{ 1 }{( x ^{-2m}+x ^{-m}+1)^{2} }) = \dfrac{x^{4m}}{2m(1+x^m+x^{2m})^2}\]

Answer 27

ya i got the correct answer..there was some net connection problem..so couldn't reply..

Answer 28

Thanks all!

Answer 29

nice idea @ganeshie8 !