Question

\[\int\limits_{0}^{oo}logx/ 1+x^{2}dx\]

will FAN and Medal

Answer 1

@Jhannybean please solve

Answer 2

@Gabylovesyou

Answer 3

\[\int_0^\infty \frac{\log(x)}{1+x^2}dx\]This?

Answer 4

can this be done by parts...

Answer 5

Or LH rule?

Answer 6

if u solve this.

Answer 7

@Jhannybean still not able to get

Answer 8

Are you just suppose to determine if it is divergent or convergent ?

Answer 9

\[x^2+1>x^2 \\ \frac{1}{x^2+1}<\frac{1}{x^2} \\ \frac{\ln(x)}{x^2+1}<\frac{\ln(x)}{x^2}\]
if so try comparision test

Answer 10

@Preetha ma'am please help!

Answer 11

can this question be solved?? @dan815 @nincompoop @Kainui please help me in this.

Answer 12

i have been trying everything but couldn't solve it

Answer 13

substitute \(x = \tan u \implies \dfrac{dx}{1+x^2} = du\), the integral becomes :

\[\int_0^\infty \frac{\ln(x)}{1+x^2}dx = \int\limits_{0}^{\pi/2} \ln(\tan u) \]

Answer 14

this comes out to be zero.

Answer 15

Next notice below to conclude the integral evaluates to 0 :
\[ \int\limits_{0}^{\pi/2} \ln(\tan u) du= \int\limits_{0}^{\pi/2} \ln(\tan (\pi/2-u))du =\int\limits_{0}^{\pi/2} \ln(\cot u)du = -\int\limits_{0}^{\pi/2} \ln(\tan u) du \]

Answer 16

\(I = -I \)

a thing is equal to its negaitve
thats only possible when it is 0

Answer 17

if this is indefinite integral?

Answer 18

then i think we simply say it is not possible to represent in elementary functions :P
http://www.wolframalpha.com/input/?i=%5Cint++%5Cfrac%7B%5Clog%28x%29%7D%7B1%2Bx%5E2%7Ddx

Answer 19

ok.