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KelumptusBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ 1 }{ 4+x ^{2} }dx=\frac{ (x\sqrt{4}i) }{ (x+\sqrt{4}i) }\] \[=\int\limits_{}^{}\frac{u2\sqrt{4}i}{u}=\int\limits_{}^{}\frac{2\sqrt{4}i}{u}=2\sqrt{4}i \lnu + c\] \[=2\sqrt{4}i \lnx+\sqrt{4}i + c\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
i didnt get your solution/or the question..please clarify..
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
The question is: is \[2\sqrt{4}i \ln x+\sqrt{4}i + c\] the indefinite integral of \[\int\limits_{}^{}\frac{1}{4+x ^{2}} dx\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
you can see 1/ 4+x^2 = 1/ (x+ 2i)(x2i) = 1/4i * (x+2i  (x2i))/(x+2i)(x2i) now separate(simplify) and integrate..hope that helped.. am sorry i dint get your soln much.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
dw:1354951402025:dw
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
That is correct also, but i am wondering if the solution involving complex numbers is also correct.
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
Actually, the solution that you typed should be 1/2 Tan(x/2) + K
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
Ahh crap, sorry, i didn't read your solution properly, i just assumed that you were using trig substitution. Forget what i just said, let me take a closer look =)
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
yes, I can show that both functions has a same power series, and it shows the equality is corect.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
for example tandw:1354952178836:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.2
for example tan^1(i)=inf
 one year ago

KelumptusBest ResponseYou've already chosen the best response.0
Ahh, thats using the power series. Cheers =)
 one year ago
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