calculate-0-logx-x-2-x-2-dx-

Question Number 138132 by mathmax by abdo last updated on 10/Apr/21

calculate \int_{0}^{\infty} \frac{logx}{x^{2} - x + 2} dx

Answered by Ñï= last updated on 13/Apr/21

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

= \frac{l n 2}{\sqrt{2}} \int_{0}^{\infty} \frac{d x}{2 x^{2} - \sqrt{2} x + 2} + \sqrt{2} \int_{0}^{\infty} \frac{l n x}{2 x^{2} - \sqrt{2} x + 2} d x = I_{1} + I_{2}

I_{2} = \sqrt{2} \int_{0}^{\infty} \frac{- l n x}{2 - \sqrt{2} x + 2 x^{2}} d x = - I_{2}

\Rightarrow I_{2} = 0

I = I_{1} = \frac{l n 2}{\sqrt{2}} \int_{0}^{\infty} \frac{d x}{{(\sqrt{2} x - \frac{1}{2})}^{2} + \frac{7}{4}} = \frac{l n 2}{2 \sqrt{\frac{7}{4}}} \tan^{- 1} \frac{\sqrt{2} x - \frac{1}{2}}{\sqrt{\frac{7}{4}}} ∣_{0}^{\infty}

= \frac{1}{\sqrt{7}} (\frac{π}{2} + \tan^{- 1} \frac{1}{\sqrt{7}}) l n 2

Facebook Tweet Pin