0-log-x-x-1-x-9-

Question Number 162513 by Lordose last updated on 30/Dec/21

\int_{0}^{\infty} \frac{\log (x)}{(x + 1) (x + 9)}

Answered by abdullahoudou last updated on 30/Dec/21

x - > \frac{9}{x}

Answered by Lordose last updated on 30/Dec/21

I \overset{x = \frac{9}{x}}{=} 9 \int_{0}^{\infty} \frac{\log (9) - \log (x)}{x^{2} (\frac{9}{x} + 1) (\frac{9}{x} + 9)} dx

I = \int_{0}^{\infty} \frac{2 \log (3) - \log (x)}{(x + 9) (x + 1)} dx

2 I = 2 \log (3) \int_{0}^{\infty} \frac{1}{(x + 9) (x + 1)} dx

2 I = \frac{\log (3)}{4} \log (\frac{x + 1}{x + 9}) ∣_{0}^{\infty}

I = \frac{\log^{2} (3)}{4}

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