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

Xét (I=intlimits_{e}^{{{e}^{2}}}{frac{f({{ln }^{2}}x)}{xln x}}dx).

Đặt (u = {ln ^2}x) ( Rightarrow du = frac{{2ln x}}{x}dx = frac{{2{{ln }^2}x}}{{xln x}}dx = frac{{2u}}{{xln x}}dx Rightarrow frac{{dx}}{{xln 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}intlimits_1^4 {frac{{f(u)}}{u}} du = frac{1}{2}intlimits_1^4 {frac{{f(x)}}{x}} dx = frac{1}{2}left( {intlimits_1^2 {frac{{f(x)}}{x}} dx + intlimits_2^4 {frac{{f(x)}}{x}} dx} right)\
= frac{1}{2}left( {intlimits_1^2 {frac{2}{{xleft( {2x – 5} right)}}} dx + intlimits_2^4 {frac{{3{x^2} + 6x}}{x}} dx} right) = frac{1}{2}left( {intlimits_1^2 {frac{2}{{xleft( {2x – 5} right)}}} dx + intlimits_2^4 {left( {3x + 6} right)} dx} right)\
= frac{1}{2}left[ {frac{4}{5}intlimits_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})