I was wondering...
"will you marry me?" right? XDD
lol
I'm writing it now, wait :D
haha okay
TYPE FASTER!!!!
lol
it gon be dramatic XD
I bet
OMFG
His computer froze :/
haha that would be funny xD
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 \[\frac{ 1 }{ 18 } \int\limits \frac{ 1 }{ \sqrt{9x^2+1} } d (9x^2+1)=\frac{ 1 }{ 18 } \ln(\sqrt{9x^2+1})+c\]
It really took you that long to write that?
It's just that openstudy doesn't use normal LaTeX operators so I can type it faster.
haha its okay. i wish i could help but this gurl is not that good in math
omg i cant do that lol
if you substitute sqrt(9x^2+1) as t then can you proceed?
@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.
As in? if you look closely in the wolfram alfa case you are having dx and here it is a function of x
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)
also, if you try to type the integral without the square root, wolframalpha uses the log formula
don't get me wrong, i can do integrals perfectly, I was just wondering why that is the case here
It's \(\Large\int\frac{du}u\) which is ln(u) instead of \(\Large\int\frac{du}{\sqrt u}\)
You can view \(\Large\int\frac{du}{\sqrt u}\) as \(\Large\int u^{-0.5} du\)
haha, yeah, got it :P
Glad you got it :)
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