Tích phân \(I = {8\ln x + 1} \) bằng:

* Hướng dẫn giải

Ta có:

\(I = \int\limits_1^e {\dfrac{{\sqrt {8\ln x + 1} }}{x}\,dx}  \)

\(= \dfrac{1}{8}\int\limits_1^e {\sqrt {8\ln x + 1} \,d\left( {8\ln x + 1} \right)}\)

\(  = \dfrac{1}{8}.\dfrac{2}{3}{\left( {8\ln x + 1} \right)^{\dfrac{3}{2}}}\left| \begin{array}{l}^e\\_1\end{array} \right.\)

\( = \dfrac{1}{{12}}\left( {{9^{\dfrac{3}{2}}} - 1} \right) = \dfrac{{13}}{6}\)