Question

find integral of (xcube-1)raiseto1/3*xraiseto5

Answer 1

( x^3 - 1) / 3 x^5

Answer 2

is this d Q?

Answer 3

how did u do it?!?

Answer 4

i didnt solve yet...i have written d integrand again...is it right?

Answer 5

exactly i dont know
...thats why im asking u how u did it

Answer 6

(xcube-1)raiseto1/3*xraiseto5 = ( x^3 - 1) / 3 x^5 ????

Answer 7

yes

Answer 8

ok

Answer 9

thts the question

Answer 10

( x^3 - 1) / 3 x^5
= 1/3 x^(-2) - 3 x^(-5)

Answer 11

1/3 x^(-2) - 1/3 x^(-5)

Answer 12

theres one more..
fine integral of 1+cosAcosx/cosA+cosx ....(A is a constant!)

Answer 13

(1+cosAcosx)/ (cosA+cosx) ???

Answer 14

1/3 is raise to that whole x^3-1

Answer 15

yes

Answer 16

(1/3) x^(-2) - (1/3 )x^(-5) but it's not d answer...u have to integrate it using integration of x^n = x^(n+1)/(n+1)

Answer 17

k..

Answer 18

is it substitution,by parts or what?!

Answer 19

\[\Large \int (x^3 - 1)^{\frac 13} x^5 dx\]
let \[u = x^3 - 1\]
\[du = 3x^2dx\]
so the integral becomes
\[\Large \implies \frac 13 \int u^{\frac 13} du \times (u-1)\]
so if you simplify...
\[\Large \implies \frac 13 \int (u^{\frac 43} - u ^{\frac 13})du\]

Answer 20

uhh wait..that should be + not -

Answer 21

(xcube-1)raiseto1/3*xraiseto5

Answer 22

wait...check d question again..

Answer 23

there is no 1/3 rd power ...

Answer 24

ill check

Answer 25

raise to 1/3 means ^1/3

Answer 26

yes

Answer 27

oh...yes...sorry

Answer 28

question is right

Answer 29

anyway...like i was saying..should be + not - in my last line \[\huge \implies \frac 13 \int (u^{\frac 43} + u^{\frac 13})du\]

Answer 30

yes...lgbasallote is right

Answer 31

now integrate that

Answer 32

k

Answer 33

thx

Answer 34

welcome