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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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.

Mathematics
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\[\int \frac{4t}{1+6t^2}\]
\[4\int \frac{t}{1+6t^2}dt \] \[u=1+6t^2\] \[du=12tdt\] \[\frac{du}{12}=tdt\]
Ah you used substitution. Thanks a lot, makes muh more sense

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Other answers:

ys and it's 1/3
wolframalpha is a good site to use to check
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 \[\int\limits \frac{\frac{1}{3} \times 12t}{1 + 6t^2} dt = \frac{1}{3} \int\limits \frac{4t}{1 + 6t^2} dt\]
the 1/8 might be a typo
Thanks to both of you :) help is greatly appreciated haha. And it looks like it. 1/3 does seem right!

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