I give up...

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
Anyone wanna try? Compute the indefinite integral:

?[ 4 x/(v{x^2 + 2})]

i got: dx = 4((x^2+2)^-(3/2)) / 3

it's wrong though. anyone?!!?

thanks
 
Dec 10, 2005
27,957
12,502
136
Just use u substitution:

u = x^2 + 2
du = 2x dx

so you get Integral of (2u^(-1/2))du

So you should get 4u^(1/2) which gives you 4sqrt(x^2 + 2) + c
 
Oct 19, 2000
17,860
4
81
I look at that, and thank the lord that I will never have to use that kind of math as long as I live.
 

xSauronx

Lifer
Jul 14, 2000
19,582
4
81
Originally posted by: blurredvision
I look at that, and thank the lord that I will never have to use that kind of math as long as I live.

qft. i could barely do high school algebra if i had to (my high school education sucked it, and i havent done algebra in almost 8 years)
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: TecHNooB
Looks like an inverse trig.. lemme try and solve this :)

Nope. Simple substitution.

If the factor of x wasn't out in front, it would be a trig substitution where you'd substitute:

cos(t) = sqrt(2)/sqrt(x^2+2). --> 1/sqrt(x^2+2) = cos(t)/sqrt(2) and go from there.
 

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
man these are all so hard for me...you guys are using methods we havent learned to solve them :(

thanks for the help!
 

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
Ok, honestly I am not looking for the answer, but why is this not right? If someone could point me in the right direction that'd be great.

solve ? 3xe^(-6x)^2

i got: e^(-6x)^2 / -4

which is wrong. ahhhhhh

 

ManyBeers

Platinum Member
Aug 30, 2004
2,519
1
81
Originally posted by: Kazaam
Ok, honestly I am not looking for the answer, but why is this not right? If someone could point me in the right direction that'd be great.

solve ? 3xe^(-6x)^2

i got: e^(-6x)^2 / -4

which is wrong. ahhhhhh

Here let me take a stab at it..............................................nope, sorry pal.
 

MovingTarget

Diamond Member
Jun 22, 2003
9,002
115
106
u substitution is probably the best method as described above. Just don't forget the "+C" constant of integration at the end.

I am a math grad student. I got a C in high school algebra. :(

Calculus may look scary, but believe me, there is a lot more to math that you might be good at. Problem is, you don't get to the fun stuff until its too late for most people. Good luck!
 

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
Thanks guys...yeah u substitution is what the homework is over. I'm having problems though, lol.
 

TecHNooB

Diamond Member
Sep 10, 2005
7,458
1
76
Originally posted by: MovingTarget
u substitution is probably the best method as described above. Just don't forget the "+C" constant of integration at the end.

I am a math grad student. I got a C in high school algebra. :(

Calculus may look scary, but believe me, there is a lot more to math that you might be good at. Problem is, you don't get to the fun stuff until its too late for most people. Good luck!

Fun stuff? Theoretical math is all gibberish I tell ya!
 

QurazyQuisp

Platinum Member
Feb 5, 2003
2,554
0
76
Originally posted by: Kazaam
Ok, honestly I am not looking for the answer, but why is this not right? If someone could point me in the right direction that'd be great.

solve ? 3xe^(-6x)^2

i got: e^(-6x)^2 / -4

which is wrong. ahhhhhh

errr... (forget the by parts)

U = 36x^2
du = 72xdx

?3(e^u)/72du
3(e^u)/72 + c => e^(36x)^2/24 +c
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: Kazaam
Ok, honestly I am not looking for the answer, but why is this not right? If someone could point me in the right direction that'd be great.

solve ? 3xe^(-6x)^2

i got: e^(-6x)^2 / -4

which is wrong. ahhhhhh

Here, you have 3x being multiplied by your exponential which is a function of x^2. Just like the previous problem, you want to use a substitution.

The form of your integral is correct, but you need to be careful with your constants.
 

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
Originally posted by: QurazyQuisp
Originally posted by: Kazaam
Ok, honestly I am not looking for the answer, but why is this not right? If someone could point me in the right direction that'd be great.

solve ? 3xe^(-6x)^2

i got: e^(-6x)^2 / -4

which is wrong. ahhhhhh

errr... (forget the by parts)

U = 36x^2
du = 72xdx

?3(e^u)/72du
3(e^u)/72 + c => e^(36x)^2/24 +c

err thats not right. how'd you get 36x? itd be -12x

im still getting (1/-4)e^(-6x)^2 which seems right, i dunno. but its not right lol. ahh
 

magomago

Lifer
Sep 28, 2002
10,973
14
76
Originally posted by: blurredvision
I look at that, and thank the lord that I will never have to use that kind of math as long as I live.

And I look at that and go "Damn...it feels good to know that I think its straight forward to solve that kind of math!" ;)
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: Kazaam
Originally posted by: QurazyQuisp
Originally posted by: Kazaam
Ok, honestly I am not looking for the answer, but why is this not right? If someone could point me in the right direction that'd be great.

solve ? 3xe^(-6x)^2

i got: e^(-6x)^2 / -4

which is wrong. ahhhhhh

errr... (forget the by parts)

U = 36x^2
du = 72xdx

?3(e^u)/72du
3(e^u)/72 + c => e^(36x)^2/24 +c

err thats not right. how'd you get 36x? itd be -12x

im still getting (1/-4)e^(-6x)^2 which seems right, i dunno. but its not right lol. ahh

Re-write the problem like this:

?3xe^(-6x)^2 dx = 3?x e^(36x^2) dx

Let u = x^2 -> du = 2xdx --> xdx = 1/2 du

Your integral becomes:

3/2 ? e^36u du

Which you should be able to solve from there.