Tích phân \(I = \int\limits_1^a {x\ln x} dx\) có g...

C

- Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{1}{x}dx\\v = \frac{{{x^2}}}{2}\end{array} \right.\).

\(\begin{array}{l} \Rightarrow I = \left. {\left( {\frac{{{x^2}}}{2}.\ln x} \right)} \right|_1^a - \int\limits_1^a {\frac{x}{2}dx} \\ = \left. {\left( {\frac{{{x^2}}}{2}.\ln x} \right)} \right|_1^a - \left. {\left( {\frac{{{x^2}}}{4}} \right)} \right|_1^a\\ = \frac{{{a^2}\ln \left| a \right|}}{2} + \frac{{1 - {a^2}}}{4}\end{array}\)