## Kelumptus Group Title Is the following a correct solution? one year ago one year ago

1. Kelumptus Group Title

$\int\limits_{}^{}\frac{ 1 }{ 4+x ^{2} }dx=\frac{ (x-\sqrt{4}i) }{ (x+\sqrt{4}i) }$ $=\int\limits_{}^{}\frac{u-2\sqrt{4}i}{u}=\int\limits_{}^{}\frac{-2\sqrt{4}i}{u}=-2\sqrt{4}i \ln|u| + c$ $=-2\sqrt{4}i \ln|x+\sqrt{4}i| + c$

2. shubhamsrg Group Title

3. Kelumptus Group Title

The question is: is $-2\sqrt{4}i \ln |x+\sqrt{4}i| + c$ the indefinite integral of $\int\limits_{}^{}\frac{1}{4+x ^{2}} dx$

4. shubhamsrg Group Title

you can see 1/ 4+x^2 = 1/ (x+ 2i)(x-2i) = 1/4i * (x+2i - (x-2i))/(x+2i)(x-2i) now separate(simplify) and integrate..hope that helped.. am sorry i dint get your soln much.

5. mahmit2012 Group Title

|dw:1354951402025:dw|

6. Kelumptus Group Title

That is correct also, but i am wondering if the solution involving complex numbers is also correct.

7. Kelumptus Group Title

Actually, the solution that you typed should be 1/2 Tan(x/2) + K

8. Kelumptus Group Title

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 =)

9. mahmit2012 Group Title

yes, I can show that both functions has a same power series, and it shows the equality is corect.

10. mahmit2012 Group Title

for example tan|dw:1354952178836:dw|

11. mahmit2012 Group Title

for example tan^-1(i)=inf

12. Kelumptus Group Title

Ahh, thats using the power series. Cheers =)