Mathematics
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OpenStudy (anonymous):
integrate sec^2 x/4+tan x
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OpenStudy (amistre64):
the top is the derivative of the bottom
OpenStudy (amistre64):
\[\int \frac{du}{u}\]
OpenStudy (anonymous):
could you show me the steps? because i cant understand it
OpenStudy (amistre64):
im not sure what you mean by steps. this is a basic antiderivative.
OpenStudy (amistre64):
its of the from:\[\int \frac{1}{u}du\]
what would that integrate to?
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OpenStudy (anonymous):
ln u ?
OpenStudy (amistre64):
yes; except u = 4+tan x
OpenStudy (amistre64):
oh, and +c on the end :)
OpenStudy (amistre64):
to dbl chk; take the derivative of ln(4+tan x) and see if you get that result
OpenStudy (anonymous):
and how about this question ? integrate x^2+1/x^3+3x ?
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OpenStudy (amistre64):
same concept; notice that the top is almost the derivative of the bottom
OpenStudy (amistre64):
what would ln(x^3+3x) derive to? and youll see whats missing
OpenStudy (anonymous):
do i write integral sign 1/u x du/3 ?
OpenStudy (amistre64):
that would work too ;)
OpenStudy (anonymous):
it would be then 1/3 x du/u right ?
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OpenStudy (amistre64):
exactly
OpenStudy (anonymous):
then i take the 1/3 out and it would be 1/u ?
OpenStudy (amistre64):
youve got it, your like a pro by now
OpenStudy (anonymous):
then i substitute it with x^3+3x ?
OpenStudy (anonymous):
thanks ! you have helped me understand it better ! ;)
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OpenStudy (amistre64):
\[\frac{1}{3}\int \frac{1}{u}du\to\ \frac{1}{3}ln(u);\ u=x^3+3x\]