Sa se calculeze
\( \int\frac{x^{2}(\ln x-1)}{x^{4}- \ln^{4} x}dx, x>1 \).
Gazeta Matematica, Olimpiada locala Constanta, 2008
Calcul de integrala 3
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Solutie.
Sa notam cu \( I \) integrala data. Prin impartire cu \( \ln^{4}x \), avem:
\( I=\int\frac{\frac{x^2}{\ln^2 x}\cdot\frac{\ln x-1}{\ln^2 x}}{\left(\frac{x}{\ln x}\right)^{4}-1}\right) dx \).
Folosind substitutia \( \frac{x}{\ln x}=t \), integrala noastra se transforma in
\( I=\frac{t^2}{t^4-1}dt=\int\left(\frac{1}{t^2-1}+\frac{1}{t^2+1}\right)dt=\frac{1}{2}\left(\frac{1}{2} \ln\frac{t-1}{t+1}+\arctan t) \).
Revenind la substitutia initiala se obtine
\( I=\frac{1}{4}\ln\frac{x-\ln x}{x+\ln x}+\frac{1}{2}\arctan\frac{x}{\ln x}+\mathcal{C} \). \( \qed \)
Sa notam cu \( I \) integrala data. Prin impartire cu \( \ln^{4}x \), avem:
\( I=\int\frac{\frac{x^2}{\ln^2 x}\cdot\frac{\ln x-1}{\ln^2 x}}{\left(\frac{x}{\ln x}\right)^{4}-1}\right) dx \).
Folosind substitutia \( \frac{x}{\ln x}=t \), integrala noastra se transforma in
\( I=\frac{t^2}{t^4-1}dt=\int\left(\frac{1}{t^2-1}+\frac{1}{t^2+1}\right)dt=\frac{1}{2}\left(\frac{1}{2} \ln\frac{t-1}{t+1}+\arctan t) \).
Revenind la substitutia initiala se obtine
\( I=\frac{1}{4}\ln\frac{x-\ln x}{x+\ln x}+\frac{1}{2}\arctan\frac{x}{\ln x}+\mathcal{C} \). \( \qed \)
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