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integral ((e^2x)-2)/((e^2x)+3)
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Use substitution.
Let u = (e^(2x) + 3) du = 2 (e^(2x) + 3) So this integral is of the form 1/2 du/u So the result is (1/2) ln ((e^2x) + 3)) + C
du = 2 (e^(2x) + 3) or only du=2e^2x
3 is constant, so derivative =0 so its just du = 2e^2x :)
ok i need the final answer
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did you try ? what did u get ?
\[\frac{ 1 }{ 2 }\ln (e^{2x}+3)-\ln (e^{2x}+3)\]
nopes, that doesn't seem correct....
yes i thought that too
you split it as e^2x/ (....) - 2/ (....) right ?
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yes
so the first part is easy, just 1/2 ln (e^2x+3) you will need partial fractions for 2nd part...
show me how to do this step (partial fractions)
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