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Mathematics 24 Online
OpenStudy (anonymous):

I was wondering...

OpenStudy (anonymous):

"will you marry me?" right? XDD

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

I'm writing it now, wait :D

OpenStudy (anonymous):

haha okay

OpenStudy (anonymous):

TYPE FASTER!!!!

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

it gon be dramatic XD

OpenStudy (anonymous):

I bet

OpenStudy (anonymous):

OMFG

OpenStudy (anonymous):

His computer froze :/

OpenStudy (anonymous):

haha that would be funny xD

OpenStudy (anonymous):

Usually \[\int\limits \frac{ 1 }{ x } dx=\ln(x)+c\] and \[\int\limits x^{n}dx=\frac{ x^{n+1} }{ n+1 }\] Then why do you get this http://www.wolframalpha.com/input/?i=integral+x%2Fsqrt%289x%5E2%2B1%29 instead of \[\frac{ 1 }{ 18 } \int\limits \frac{ 1 }{ \sqrt{9x^2+1} } d (9x^2+1)=\frac{ 1 }{ 18 } \ln(\sqrt{9x^2+1})+c\]

OpenStudy (anonymous):

It really took you that long to write that?

OpenStudy (anonymous):

It's just that openstudy doesn't use normal LaTeX operators so I can type it faster.

OpenStudy (anonymous):

haha its okay. i wish i could help but this gurl is not that good in math

OpenStudy (squirrels):

omg i cant do that lol

OpenStudy (ankit042):

if you substitute sqrt(9x^2+1) as t then can you proceed?

OpenStudy (anonymous):

@ankit042 I can solve that integral, I was just wondering why the thing that I get with one of the fomulas is not equal to the other.

OpenStudy (ankit042):

As in? if you look closely in the wolfram alfa case you are having dx and here it is a function of x

OpenStudy (anonymous):

Yeah, but it's basically the same, look at \[\frac{ 1 }{ 18 } \int\limits \frac{ 1 }{ \sqrt{9x^2+1} } d (9x^2+1) \] and apply the two formulas. You should be able to apply them both. Wolframalpha gives the answer as just the second formula. Also, when i try to equate them, there are no roots (i.e. they are nowhere near as equal)

OpenStudy (anonymous):

also, if you try to type the integral without the square root, wolframalpha uses the log formula

OpenStudy (anonymous):

don't get me wrong, i can do integrals perfectly, I was just wondering why that is the case here

OpenStudy (kc_kennylau):

It's \(\Large\int\frac{du}u\) which is ln(u) instead of \(\Large\int\frac{du}{\sqrt u}\)

OpenStudy (kc_kennylau):

You can view \(\Large\int\frac{du}{\sqrt u}\) as \(\Large\int u^{-0.5} du\)

OpenStudy (anonymous):

haha, yeah, got it :P

OpenStudy (kc_kennylau):

Glad you got it :)

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