Đáp án:
$I = \ln3$
Giải thích các bước giải:
\(\begin{array}{l}
\quad I = \displaystyle\int\limits_1^3\dfrac{(x-1)^2}{x}dx\\
\to I = \displaystyle\int\limits_1^3\dfrac{x^2 - 2x + 1}{x}dx\\
\to I =\displaystyle\int\limits_1^3\left(x - 2 + \dfrac1x\right)dx\\
\to I = \left(\dfrac{x^2}{2} - 2x + \ln|x|\right)\Bigg|_1^3\\
\to I = \left(\dfrac{3^2}{2} - 2.3 + \ln3\right) - \left(\dfrac{1^2}{2} - 2.1 + \ln1\right)\\
\to I = \ln3
\end{array}\)
