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how do you do the following integral :
int (4t / (1+6t^2))
I know the answer is 1/8 ln ( 1 +6t^2) but in the process, the answer shows the numerator becoming 32t and I have no idea where the 1/8 came from.
 one year ago
 one year ago
how do you do the following integral : int (4t / (1+6t^2)) I know the answer is 1/8 ln ( 1 +6t^2) but in the process, the answer shows the numerator becoming 32t and I have no idea where the 1/8 came from.
 one year ago
 one year ago

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Outkast3r09Best ResponseYou've already chosen the best response.2
\[\int \frac{4t}{1+6t^2}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.2
\[4\int \frac{t}{1+6t^2}dt \] \[u=1+6t^2\] \[du=12tdt\] \[\frac{du}{12}=tdt\]
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Ah you used substitution. Thanks a lot, makes muh more sense
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.2
wolframalpha is a good site to use to check
 one year ago

campbell_stBest ResponseYou've already chosen the best response.0
ok... so it looks like a log function to start s = 1 + 6u^2 ds/dt = 12t but you only need 4t so it becomes 1/3 of ds/dt so \[\int\limits \frac{\frac{1}{3} \times 12t}{1 + 6t^2} dt = \frac{1}{3} \int\limits \frac{4t}{1 + 6t^2} dt\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.2
the 1/8 might be a typo
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Thanks to both of you :) help is greatly appreciated haha. And it looks like it. 1/3 does seem right!
 one year ago
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