integrate:

[(1/3)x^3]*(1+x^4)^(1/2)

u = 1 + x^4

du = 4x^3dx

but i have no idea how to finish it from here any help would be greatly appreciated...i took cal 1 about a year ago and i forgot how to properly use u substitution

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- Jan 30th 2011, 01:31 PMmaybnxtseasnu substitution problem
integrate:

[(1/3)x^3]*(1+x^4)^(1/2)

u = 1 + x^4

du = 4x^3dx

but i have no idea how to finish it from here any help would be greatly appreciated...i took cal 1 about a year ago and i forgot how to properly use u substitution - Jan 30th 2011, 01:46 PMskeeter
$\displaystyle \displaystyle \frac{1}{3} \int \frac{x^3}{\sqrt{1+x^4}} \, dx

$

$\displaystyle u = 1 + x^4$

$\displaystyle du = 4x^3 \, dx$

$\displaystyle \displaystyle \frac{1}{12} \int \frac{4x^3}{\sqrt{1+x^4}} \, dx$

$\displaystyle \displaystyle \frac{1}{12} \int \frac{1}{\sqrt{u}} \, du$

can you finish? - Jan 30th 2011, 02:03 PMmaybnxtseasn
how did u get x^3 in the numerator? did u missread my problem?

- Jan 30th 2011, 02:20 PMArchie Meade
$\displaystyle \displaystyle\frac{1}{3}\int{x^3\sqrt{1+x^4}}\;dx$

$\displaystyle \displaystyle\ u=1+x^4\Rightarrow\frac{du}{dx}=4x^3\Rightarrow\fr ac{du}{4}=x^3dx$

$\displaystyle \displaystyle\int{x^3\sqrt{1+x^4}\;dx=\int{\sqrt{1 +x^4}\;x^3dx$

The integral becomes

$\displaystyle \displaystyle\frac{1}{12}\int{u^{\frac{1}{2}}du$ - Jan 30th 2011, 02:25 PMskeeter
I did ... should be $\displaystyle \displaystyle \frac{1}{3} \int x^3 \sqrt{1+x^4} \, dx$

same sub ...

$\displaystyle \displaystyle \frac{1}{12} \int 4x^3 \sqrt{1+x^4} \, dx$

$\displaystyle \displaystyle \frac{1}{12} \int \sqrt{u} \, du$

thought I saw a negative on the 1/2 exponent. - Jan 30th 2011, 02:40 PMmaybnxtseasn
ive worked my problem down to the last step both of u have...now do i just integrate the simple equation with u? and then plug in for u and du? to get my final awnser?

so for example

if i integrate your last equation i get

(2/3)u^(3/2) du <---- do i just plug in (1+x^4) for u and im done? - Jan 30th 2011, 02:49 PMArchie Meade
Yes, as your final solution needs to be in terms of x.

Also add on the constant of integration.

The u-substitution allows a "convenient" solution.

For the calculation of definate integrals (not necessary here), you may choose to return to x,

or evaluate the definate integral using u with a change of limits to u. - Jan 30th 2011, 05:07 PMmaybnxtseasn
(1/18)(1+x^4)^(3/2) du <----- do i just change du to dx or am i missing something for the final answer because that seems to easy?

- Jan 30th 2011, 05:28 PMskeeter
- Jan 31st 2011, 01:15 AMTheCoffeeMachine
$\displaystyle \displaystyle \begin{aligned} \frac{1}{3}\int{x^3\left(1+x^4}\right)^{\frac{1}{2 }}\;{dx} & = \frac{1}{4(3)}\int{\left(1+x^4}\right)^{\frac{1}{2 }}(1+x^4)'\;{dx} \\& = \frac{2}{3(4)(3)}\int{\left[\left(1+x^4}\right)^{\frac{3}{2}}\right]'\;{dx} \\& = \frac{1}{18}\left(1+x^4}\right)^{\frac{3}{2}}+k.\e nd{aligned}$

- Jan 31st 2011, 03:16 AMtom@ballooncalculus
@ maybenxtseasn: The Coffee Machine's point relates (I think) to your earlier question about u-substitution and the chain rule... to which the answer was yes, the first one is only for navigating the second. So if the first is bogging down you might want to try a different navigational aid (if only to see how the u-sub should be working). Here's my version of what TCM is pointing out...

http://www.ballooncalculus.org/draw/...n/fourteen.png

To find F, integrate with respect to the dashed balloon (which is like where you integrate with respect to u)...

http://www.ballooncalculus.org/draw/.../fourteena.png

So you have...

http://www.ballooncalculus.org/draw/.../fourteenb.png

... where (key in spoiler) ...

__Spoiler__:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jan 31st 2011, 03:53 AMArchie Meade
The purpose of the u-substitution was to make it easy!

$\displaystyle \displaystyle\frac{d}{dx}\left(x^4+C\right)=4x^3\R ightarrow\ \int{d\left(x^4+C\right)}=\int{4x^3}\;dx\Rightarro w\int{4x^3}dx=x^4+C$

$\displaystyle \displaystyle\frac{d}{du}\left(u^4+C\right)=4u^3\R ightarrow\int{4u^3}du=u^4+C$

$\displaystyle \displaystyle\int{\sqrt{1+x^4}\;x^3dx=\int{\sqrt{u }}\;du$

when $\displaystyle u=1+x^4$