Cho hàm số . Tích phân \(I=\int\limits_{e}^{{{e}^{2}}}{\frac{f({{\ln }^{2}}x)}{x\ln x}}dx\) bằng

* Hướng dẫn giải

Xét \(I=\int\limits_{e}^{{{e}^{2}}}{\frac{f({{\ln }^{2}}x)}{x\ln x}}dx\).

Đặt \(u = {\ln ^2}x\) \( \Rightarrow du = \frac{{2\ln x}}{x}dx = \frac{{2{{\ln }^2}x}}{{x\ln x}}dx = \frac{{2u}}{{x\ln x}}dx \Rightarrow \frac{{dx}}{{x\ln x}} = \frac{{du}}{{2u}}.\)

Đổi cận : \(\left\{ \begin{array}{l} x = e \Rightarrow u = 1\\ x = {e^2} \Rightarrow u = 4 \end{array} \right.\).

Khi đó

\(\begin{array}{l} I = \frac{1}{2}\int\limits_1^4 {\frac{{f(u)}}{u}} du = \frac{1}{2}\int\limits_1^4 {\frac{{f(x)}}{x}} dx = \frac{1}{2}\left( {\int\limits_1^2 {\frac{{f(x)}}{x}} dx + \int\limits_2^4 {\frac{{f(x)}}{x}} dx} \right)\\ = \frac{1}{2}\left( {\int\limits_1^2 {\frac{2}{{x\left( {2x - 5} \right)}}} dx + \int\limits_2^4 {\frac{{3{x^2} + 6x}}{x}} dx} \right) = \frac{1}{2}\left( {\int\limits_1^2 {\frac{2}{{x\left( {2x - 5} \right)}}} dx + \int\limits_2^4 {\left( {3x + 6} \right)} dx} \right)\\ = \frac{1}{2}\left[ {\frac{4}{5}\int\limits_1^2 {\left( {\frac{1}{{2x - 5}} - \frac{1}{{2x}}} \right)} dx + \left. {\left( {\frac{{3{x^2}}}{2} + 6x} \right)} \right|_2^4} \right] = \frac{1}{2}\left[ {\frac{4}{5}.\frac{1}{2}\left. {\ln \left| {\frac{{2x - 5}}{{2x}}} \right|} \right|_1^2 + 30} \right]\\ = \frac{1}{2}\left[ {\frac{2}{5}\left( { - \ln 6} \right) + 30} \right] = 15 - \frac{1}{5}\ln 6 \end{array}\)