Question Number 11420 by ainstain last updated on 25/Mar/17

please ∫_0 ^∞ ((xlogx)/((1+x^2 )^2 ))dx=

Commented byFilupS last updated on 26/Mar/17

is log(x) in base 10 or base e?

Answered by sm3l2996 last updated on 25/Mar/17

{_(v′=(x/((1+x^2 )^2 ))) ^(u=log(x)) ⇒{_(v=((−1)/(2(1+x^2 )))) ^(u^′ =(1/x)) I=∫_0 ^∞ ((xlog(x))/((1+x^2 )^2 ))dx=[((−log(x))/(2(1+x^2 )))]_0 ^∞ +∫_0 ^∞ (dx/(2x(1+x^2 ))) I=lim_(x→∞) (−((log(x))/(2(1+x^2 ))))−lim_(x→0) (−((log(x))/(2(1+x^2 ))))+(1/2)∫_0 ^∞ (dx/(x(1+x^2 ))) (1/(x(1+x^2 )))=(a/x)+((bx+c)/(1+x^2 )) a=1; {_(((c−b)/2)−1=((−1)/2)) ^(((b+c)/2)+1=(1/2)) ⇔{_(c−b=1) ^(b+c=−1) b=−1; c=0 so: (1/(x(1+x^2 )))=(1/x)−(x/(1+x^2 )) ∫_0 ^∞ (dx/(x(1+x^2 )))=∫_0 ^∞ (dx/x)−∫_0 ^∞ ((xdx)/(1+x^2 )) =[log∣x∣−(1/2)log∣1+x^2 ∣]_0 ^∞ =lim_(x→+∞) log((x/(√(1+x^2 ))))−lim_(x→0) log((x/(√(1+x^2 )))) I=lim_(x→+∞) log((x/(√(1+x^2 ))))−lim_(x→0) (log((x/(√(1+x^2 ))))+((log(x))/(2(1+x^2 ))))

Commented bysm3l2996 last updated on 25/Mar/17

working

Answered by ajfour last updated on 26/Mar/17

zero