Yo necesito CALCULUS HELP

[GoldenBear]

How do I integrate

x * sqrt (x+1)

It seems simple enough but I can't seem to do it!

Or, if you could find a way to integrate

x^5 * sqrt (x^3+1)

that would be great too.

What I did was set u = x^3, du = 3x^2 and got u * sqrt (u+1) du

Test tomorrow

 

[shikhan]

<< How do I integrate x * sqrt (x+1) It seems simple enough but I can't seem to do it! Or, if you could find a way to integrate x^5 * sqrt (x^3+1) that would be great too. What I did was set u = x^3, du = 3x^2 and got u * sqrt (u+1) du Test tomorrow >>



Haven't done this in a while but isn't it:

Integral( x^5 * sqrt(x^3 +1) ) dx
u = X^3 +1
du = 3x^2 dx
dx = du / 3x^2

(1/3)(u-1)sqrt(u) du

= [u ^(3/2)]/3 - u ^(1/2) du


and thats not to bad from there

 

[GoldenBear]

HOLY CRAP YOU'RE A MOTHER F-EN GENIUS AND I LOVE YOU.

 

[shikhan]

a tip my calculus teacher told me...

In general, when doing u substitution, allways try to fit the constants into u if possible... Ie, you did u = x^3, where you should have done u = X^3 +1.... because du is the same either way


Good luck on your test

 

[RossGr]

This may be a bit late for you but....

let u= x+1 so du =dx

x((x+1)^1/2=> (u-1)(u^1/2) = (u^3/2 - u^1/2)du This is easy now!

In the original problem let u = x^3 + 1 and it will work better also!

Good luck on the test!

Edit:
Dam I need to type faster!

 

[GoldenBear]

<< Good luck on your test >>



<< Good luck on your test! >>

You guys are too kind

 

[Logix]

Wait. Does the chain rule work on that sqrt function? Or do you HAVE to do a U-sub?

 

[GoldenBear]

<< Wait. Does the chain rule work on that sqrt function? Or do you HAVE to do a U-sub? >>

Chain rule is only for derivates I'm sure..

 

[shikhan]

<<

<< Wait. Does the chain rule work on that sqrt function? Or do you HAVE to do a U-sub? >>

Chain rule is only for derivates I'm sure.. >>



Yup yup... u-sub is what you use your doing Integrals, chain rule for deriv's

 

[ikar0s]

Ugh, I despise integration by parts. I did learn something cool in vector calc. class a couple days back...the tabular method. Works pretty well, especially on integrations of the 3rd order or beyond.

 

[thEnEuRoMancER]

The easiest way is to integrate by parts

int(u*dv) = u*v - int(v*du)

in our case

u = x............du = dx
dv = sqrt(x+1)...........v = 2/3 * (x+1)^(3/2)

so

int (x * sqrt (x+1)) = x*2/3 * (x+1)^(3/2) - int(2/3 * (x+1)^(3/2)*dx) = x*2/3 * (x+1)^(3/2) - (2/3)*(2/5)*(x+1)^(5/2)

 

[Shaka]

udv uv vdu !!! Integration by parts is the best.

 

[Aelus]

check out this method for irrational functions:

int(x*sqrt(x^2+1)dx)=

t^2=x+1
2tdt=dx
x=t^2-1

int((t^2-1)*t*2tdt)=

split up into 2 integrals, then finish the integration, and resubstitute.

this is the easiest way to work with roots, just put t to the inverse of the power of the root.

Aelus