Value of the given integral

Find out the value of the integral

鈭?from 0 to 鈭? (ln(x)/(x^2+1)^2)dx

I know the value of lnx/(x^2+1) gives zero(but not sure about lnx/(x^2+1)^2). And the denominator somewhat reminds me of tanx. But I cannot make ends meet and figure out the answer. So any help would be deeply appreciated. If somebody can give a proper step by step solution(at least the first few steps with proper substitution) I would be much obliged.

Thanks a lot in advance :-)

J = 鈭?[x=0,鈭瀅 ln(x)dx/(x虏+1)虏

Let x=1/u with dx=鈭抎u/u虏

J = 鈭?[u=鈭?0] 鈭抣n(u)(鈭抎u/u虏) / (1/u虏+1)虏 = 鈭掆埆 [u=0,鈭瀅 u虏ln(u)du / (1+u虏)虏

鈭扟 = 鈭?[u=0,鈭瀅 uln(u) udu/(1+u虏)虏

鈭?by parts : 鈭扟 = (鈭捖?(1+u虏) * uln(u)), [u=0,鈭瀅 + 陆 鈭?[u=0,鈭瀅 {1/(u虏+1)}{ln(u)+1}du

Because lim [u鈫?] uln(u)/(1+u虏) = lim [u鈫掆垶] uln(u)/(1+u虏) = 0

鈭扟 = 陆 鈭?[u=0,鈭瀅 ln(u)du/(u虏+1) + 陆 鈭?[u=0,鈭瀅 du/(u虏+1) = 0 + 陆tan鈦宦?u), [u=0,鈭瀅

鈭扟 = 蟺/4 or J=鈭捪€/4



鈭?[u=0,鈭瀅 ln(u)du/(u虏+1) = 0 is easily confirmed with the same sub