advanced-integral-Evaluate-I-0-4xln-x-x-4-2x-2-4-dx-m-n-1970-

Question Number 117574 by mnjuly1970 last updated on 12/Oct/20

\dots a d v a n c e d i n t e g r a l \dots

E v a l u a t e ::

I := \int_{0}^{\infty} \frac{4 x l n (x)}{x^{4} + 2 x^{2} + 4} d x = ? ?

\dots m . n . 1970 . .

Answered by mindispower last updated on 12/Oct/20

x^{2} = t \Rightarrow

I = \int_{0}^{\infty} \frac{l n (t)}{t^{2} + 2 t + 4}

t \Rightarrow 2 s \Rightarrow \int_{0}^{\infty} \frac{l n (2 s) . 2 d s}{4 (s^{2} + s + 1)} = \int_{0}^{\infty} \frac{l n (2) d s}{2 (s^{2} + s + 1) = I} + \frac{1}{2} \int_{0}^{\infty} \frac{l n (x)}{x^{2} + x + 1 = J}

f i r s t I b a s i c

J = \frac{1}{2} \int_{0}^{1} \frac{l n (x)}{x^{2} + x + 1} d x + \frac{1}{2} \int_{1}^{\infty} \frac{l n (x)}{x^{2} + x + 1} d x

x = \frac{1}{t} i n 2 n d \Rightarrow d x = \frac{- d t}{t^{2}}

= \frac{1}{2} \int_{1}^{0} \frac{l n (t)}{1 + t + t^{2}} = - \frac{1}{2} \int_{0}^{1} \frac{l n (x)}{1 + x + x^{2}}

\Rightarrow w e h a v e j u s t e t o f i n d

\int_{0}^{\infty} \frac{l n (2) d x}{x^{2} + x + 1} . \frac{1}{2} = \frac{l n (2)}{2} \int_{0}^{\infty} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}

= \frac{2 l n (2)}{3} \int_{0}^{\infty} \frac{d x}{(\frac{2 x}{\sqrt{3}} + \frac{1}{\sqrt{3}})} = \frac{l n (2)}{\sqrt{3}} {[a r c t a n (\frac{2 x}{\sqrt{3}} + \frac{1}{\sqrt{3}})]}_{0}^{\infty}

= \frac{l n (2)}{\sqrt{3}} [\frac{π}{2} - \frac{π}{6}] = \frac{π l n (2)}{3 \sqrt{3}}

Commented by mnjuly1970 last updated on 12/Oct/20

t a y e b a l l a h t h a n k y o u s o

m u c h \dots

Commented by mindispower last updated on 14/Oct/20

w i t h e ? p l e a s u r