Biết rằng \(\int\limits_1^a {\ln xdx = 1 + 2a,\left( {a > 1} \right)} \).

* Hướng dẫn giải

Ta có: \(\int\limits_1^a {\ln xdx = 1 + 2a\left( {a > 1} \right)} \) 

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

\(\begin{array}{l}
 \Rightarrow I = x\ln x\left| \begin{array}{l}
^a\\
_1
\end{array} \right. - \int\limits_1^a {dx = a\ln a - x} \left| \begin{array}{l}
^a\\
_1
\end{array} \right. = a\ln a - a + 1\\
 \Rightarrow 1 + 2a = a\ln a - a + 1 \Leftrightarrow 3a = a\ln a \Leftrightarrow \ln a = 3 \Leftrightarrow a = {e^3} \approx 20,08 \in \left( {18;21} \right)
\end{array}\)