Question

For this intgeral could i let u=x^3+1, then let x=3root (u-1)

Answer 1

What integral?

Answer 2

\[\int\limits_{}^{} (x^5)(\sqrt[4]{x^3+1})\]

Answer 3

wait a sec

Answer 4

Satellite, if you want to take a stab at it you can

Answer 5

then
\[u-1=x^3\] so
\[x^5=(u-1)^{\frac{5}{3}}\] does that help?

Answer 6

oh no i see split
\[x^5=x^2\times x^3\] maybe that will work

Answer 7

I attempted that approach

Answer 8

the splitting approach

Answer 9

before i cheat let me try that
\[\int \sqrt[4]{x^3+1}x^2x^3dx\]
\[

Answer 10

\[u=x^3+1\]
\[du=3x^2\]
\[x^3=u-1\] should work right?

Answer 11

let me try

Answer 12

get
\[\frac{1}{3}\int \sqrt[4]{u}(u-1)du\]

Answer 13

Looks good

Answer 14

oh sorry i wrote before i saw your post.

Answer 15

I was on the right track. No I hadn't posted anything yet

Answer 16

it cool let me see if i can replicate

Answer 17

okay i got: (4/27)(x^3+1)^9/4 -(4/5)(x^3+1)^5/4

Answer 18

?

Answer 19

please check!!