Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-

!!! CALCULUS!!! ><

Name: Anonymous 2007-12-06 19:20

lol coming to 4chan... the internet hate machine is definately my last resort... but please some one HELP ME WITH THIS CALCULUS PROBLEMMM AHHHHHH

THE POWER LEVEL OF THIS EQUATION IS OVER 9000!!!!

I CAN'T FIND THE INDEFINITE INTEGRAL OF THIS EQ.

         ...(1-(x)^(1/2))/(1+x^(1/2))

I tried substitution but my equation got only more complicated....

some anonymous mathmeticians/scientists.... one please help... thanks

Name: Anonymous 2007-12-06 19:23

the internet hate machine
OVER 9000
!!!!
This is not /b/.
Ask as if you weren't a brain-damaged moron and you might get some help.

Name: Anonymous 2007-12-06 19:28

(1-(x)^(1/2))/(1+x^(1/2))
(-x-1)/(x-1)
done in 30 seconds you fucking moron

Name: Anonymous 2007-12-06 19:37

I doubt you >>3

Name: Anonymous 2007-12-06 19:51

>>3
I am asking for the indefinate integral... you have failed. >:[[]]

okay anyways seriously...

I have managed to progress further by manipulating the substitution equation...
u=1 + x^(1/2)
and resulted with
u - 1 = x^(1/2)

When I sub in u-1 for x^(1/2) 
I get (2 - u) / u

Integrating this I get 2ln(u) - u and even after substituting u =
1+x^(1/2) I still don't get the text book's answer

 (5 + 4(x)^(1/2) - x - 4ln(1+ (x)^(1/2)) + C) but this instead...

2ln(1+ (x)^(1/2)) + 1+ (x)^(1/2)

Please help me anonymous mathmeticians/scientists

Name: Anonymous 2007-12-06 19:55

>>5
1+x^(1/2) != 0
x^(1/2) != -1
x^(1/2)^2 != (-1)^2
x != (-1)^2
x != 1

or from what I answered earlier

x-1 != 0
x != 1

Name: Anonymous 2007-12-06 20:16

sorry I thought indefinite integral was something else, but still easy
y=(-x-1)/(x-1)
y=(-2)/(x-1)-1
y-1=(-2)/(x-1)
(y-1)/(-2) = 1/(x-1)
(-2)/(y-1) = x-1

Name: Anonymous 2007-12-06 20:17

(-2)/(y-1)+1 = x

Name: Anonymous 2007-12-06 20:32

u = 1 + sqrt(x)
du = 1/[2sqrt(x)]dx
sqrt(x) = u - 1, 2sqrt(x) = 2u - 2

du = 1/(2u - 2) dx
dx = (2u - 2) du

so the integral becomes

[1 - sqrt(x)]/u * (2u - u) du

1 - sqrt(x) = -u + 2

(-u + 2) (2u - u) / u du

FOIL the top, divide each term by u, integrate each term seperately

Name: Anonymous 2007-12-06 20:36

That last part should be:


[1 - sqrt(x)]/u * (2u - 2) du

1 - sqrt(x) = -u + 2

(-u + 2) (2u - 2) / u du

Name: Anonymous 2007-12-06 21:06

>>7 &
>>8

No idea what you're doing...

>>9
Thats exactly right...
I got the answer! sweet thanks anonymous mathmeticians/scientists....

EPIC WIN!

BTW is 10 ... i think you didn't see 9... lol

Again thanks :)

Name: Anonymous 2007-12-06 22:17

(%i1) integrate((1-(x)^(1/2))/(1+x^(1/2)), x);
(%o1) 2*(-(x-4*sqrt(x))/2-2*log(sqrt(x)+1))
(%i2) ratsimp(%);
(%o2) -x+4*sqrt(x)-4*log(sqrt(x)+1)

Don't change these.
Name: Email:
Entire Thread Thread List