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

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0i didnt get your solution/or the question..please clarify..

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

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0you 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.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1354951402025:dw

Kelumptus
 2 years ago
Best ResponseYou've already chosen the best response.0That is correct also, but i am wondering if the solution involving complex numbers is also correct.

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

Kelumptus
 2 years ago
Best ResponseYou've already chosen the best response.0Ahh 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 =)

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.2yes, I can show that both functions has a same power series, and it shows the equality is corect.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.2for example tandw:1354952178836:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.2for example tan^1(i)=inf

Kelumptus
 2 years ago
Best ResponseYou've already chosen the best response.0Ahh, thats using the power series. Cheers =)
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