Integration using Table of integrals? Integrate ( e^x ) dx / (3-e^2x) I'm stumped...
\[ \large \int\frac{e^x}{3-e^{2x}}\,dx \] ?
yep.
on my table of integrals, I thought maybe using the int du/ u sqrt( a +bu) would work but idk how to set it up
u already learned substitution, right?
u sub yes... but im not sure for this one...
we have to use the table of integrals...
u use substitution in order to turn an integral that is not on the table into one that IS on the table.
u can rewrite the integral into \[ \large \int\frac{e^x}{3-(e^x)^2}\,dx \] right?
yes
so u would be e^x
du is e^x as well?
yes
so if u=e^x then du=e^x dx therefore the integral becomes \[ \large \int\frac{1}{3-u^2}\,du \] agree?
this one should be on your table. right?
im looking...
i have it. do u want me to post it?
please
theres too many agh!
no problem \[ \large \color{blue}{\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C} \] for \(a\neq0\)
thank you so much!!!!
compare with what we got \[ \large \int\frac{1}{3-u^2}\,du \] a^2=.....
did you get this from a table? what number was it? i cant find this one in my table... but its definetly the right one i just cant find it
we must have different tables. i have a russian book with me.
what is the value of a^2 ?
i see... ill keep looking..
found it!
I have another question regarding these integrals using the tables. could you help?
ill post a new question so you get credit
sure. but, shouldn't we first finish this one?
i finished it i just needed to find the formula from the tables... a is sqrt 3 and then you simply plug the rest in right?
and replace the value of u=e^x
yes.
ok.
i will close this and open a new question..
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