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anonymous
 3 years 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.
anonymous
 3 years 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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{4t}{1+6t^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[4\int \frac{t}{1+6t^2}dt \] \[u=1+6t^2\] \[du=12tdt\] \[\frac{du}{12}=tdt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah you used substitution. Thanks a lot, makes muh more sense

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wolframalpha is a good site to use to check

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.0ok... 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 \[\int\limits \frac{\frac{1}{3} \times 12t}{1 + 6t^2} dt = \frac{1}{3} \int\limits \frac{4t}{1 + 6t^2} dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 1/8 might be a typo

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