Question

what is the antiderivative of (x^2)/(9 + x^(6))

Answer 1

\[\int \frac{x^2}{9+x^6}dx\]

Answer 2

yes

Answer 3

You need to apply integration by parts twice to the function \[x^2(9+x^6)^{-1}\]

Answer 4

that would work out great :)

Answer 5

Or maybe not (how to find integral of 1/9+x^6?

Answer 6

..... if we could do that

Answer 7

\[(x^2)^3\]
comes to mind, but may not be helpful

Answer 8

Substitute y = x^3... it should work I think!

Answer 9

\[(3x^3)^2\]would be nice maybe

Answer 10

use the substitution u=x^3?

gives you arctan(x^3/3)/9??

Answer 11

Possible intermediate steps:
integral x^2/(9+x^6) dx
For the integrand x^2/(x^6+9), substitute u = x^3 and du = 3 x^2 dx:
= 1/3 integral 1/(u^2+9) du
The integral of 1/(u^2+9) is 1/3 tan^(-1)(u/3):
= 1/9 tan^(-1)(u/3)+constant
Substitute back for u = x^3:
= 1/9 tan^(-1)(x^3/3)+constant

Answer 12

Let y = x^3
dy/dx = 3x^2
dy/3 = x^2 dx
\[\int{\frac{x^2}{9+x^6}}dx=1/3\int{\frac{1}{9+y^2}}dy\] which can be evaluated

Answer 13

straight off the wolfram lol

Answer 14

yeh hate that dont you amistre

Answer 15

oh i like it when i get stuck, but to claim it as your own is a bit misleading ;)

Answer 16

yeah, substituting y=x^3 works a treat :)

Answer 17

yeh thats what i mean haha its a godsend when your stuck

Answer 18

Wolfram gives us:

Possible intermediate steps:
integral x^2/(9+x^6) dx
For the integrand x^2/(x^6+9), substitute u = x^3 and du = 3 x^2 dx:
= 1/3 integral 1/(u^2+9) du
The integral of 1/(u^2+9) is 1/3 tan^(-1)(u/3):
= 1/9 tan^(-1)(u/3)+constant
Substitute back for u = x^3:
= 1/9 tan^(-1)(x^3/3)+constant

Answer 19

I got the answer already guys :)

Answer 20

so then from there xactxx where do i go?

Answer 21

... on to the next problem :)

Answer 22

hahahaah good answer :)

Answer 23

arctangent yes?

Answer 24

yess

Answer 25

yessss