Question

how do you integrate x^2(rad(3x+4))

Answer 1

\[\int\limits x^2/\sqrt{3x+4}\] ?

Answer 2

your rad(3x+4) function makes no sense

Answer 3

\[\int\limits x^{2}\sqrt{3x+4}\]

Answer 4

Oh its multiplied.

Answer 5

sorry for confusion

Answer 6

Use substitution for the \[\sqrt{3x+4}\]

Answer 7

What I did was: U=(3x+4)^(1/2) then solve for x. I got x=(1/3)(u^2-4). Find dx. dx=(1/3)(2u). Then plug it all in:
\[\int\limits((1/3)u^2-4/3)(u)(2u/3)du=(2/9)\int\limits(u^2-4)(u^2)du.\]
Straight forward from there. :)

Answer 8

Miscopy. That should read:\[\int\limits((1/3)u^2-4/3)^2(u)(2u/3)du\].
That what you did myininaya?

Answer 9

\[u=3x+4, du=3 dx\]
\[x=\frac{u-4}{3}\]
\[x^2=\frac{(u-4)^2}{9}\]
\[\int\limits_{}^{} \frac{1}{3}*\frac{(u-4)^2}{9}*\sqrt{u} du\]
\[\frac{1}{27}\int\limits_{}^{}(u^2-8u+16)*\sqrt{u} du\]
\[\frac{1}{27}\int\limits_{}^{}(u^{\frac{5}{2}}-8u^{\frac{3}{2}}+16u) du\]

Answer 10

That works too :P

Answer 11

Mine was a tad more complicated looking haha

Answer 12

where is the symbol for nice fractions like that myininaya?

Answer 13

\[\frac{1}{27}*(\frac{2}{7}u^\frac{7}{2}-8*\frac{2}{5}u^\frac{5}{2}+16*\frac{1}{2}u^2)+C\]

Answer 14

use frac{2}{3} will give you 2/3 in the equation editor

Answer 15

nice thanks

Answer 16

\[\frac{7}{x}\] I like it!!!

Answer 17

we need to put that junk in terms of x

Answer 18

yea lol make him do it I say

Answer 19

Agreed^^

Answer 20

i agree too :)