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i am trying to integrate
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I am writing it now
\[\int\limits_{}^{} (3\sec^2t)/(6+3tant) dt\]
can me move something out as a constant?
You can factor out a 3.
only the 3 right?
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the 6+3 on the bottom stays right?
i integrated and got the right answer but there is a constant 3 still left
factor a 3 out of the top and bottom perhaps ...
so would i be left with 2+1 on the bottom??
the top looks to be a derivative of the bottom if i see it right
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hmm so its in that form du/u
6 + 3tan(t) derives to 0 + 3sec^2 right?
so take that bottom as u
that is of the form du/u if im right :)
thanks
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youre welcome :)
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