Đáp án:
$I = 16 + \dfrac12\ln5$
Giải thích các bước giải:
\(\begin{array}{l}
\quad I = \displaystyle\int\limits_1^3\left(4x + \dfrac{1}{2x-1}\right)dx\\
\to I = \displaystyle\int\limits_1^34xdx + \displaystyle\int\limits_1^3\dfrac{1}{2x-1}dx\\
\to I = 4\displaystyle\int\limits_1^3xdx + \dfrac{1}{2}\displaystyle\int\limits_1^3\dfrac{1}{2x-1}d(2x-1)\\
\to I = 2x^2\Bigg|_1^3 + \dfrac{1}{2}\ln|2x-1|\Bigg|_1^3\\
\to I = 2(3^2 - 1^2) + \dfrac12\left(\ln5 - \ln1\right)\\
\to I = 16 + \dfrac12\ln5
\end{array}\)
