Đáp án:
\(I = \dfrac{3^{2018} - 2^{2018}}{4036}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad I = \displaystyle\int\limits_1^2\dfrac{(x+2)^{2017}}{x^{2019}}dx\\
\Leftrightarrow I = -\dfrac12\displaystyle\int\limits_1^2\left(\dfrac{x+2}{x}\right)^{2017}\cdot \left(-\dfrac{2}{x^2}\right)dx\\
\text{Đặt}\ u = \dfrac{x+2}{x}\\
\Rightarrow du = - \dfrac{2}{x^2}dx\\
\text{Đổi cận:}\\
\begin{array}{c|ccc}x&1&&&2\\\hline u&3&&&2\end{array}\\
\text{Ta được:}\\
\quad I = - \dfrac12\displaystyle\int\limits_3^2u^{2017}du\\
\Leftrightarrow I = \dfrac12\displaystyle\int\limits_2^3u^{2017}du\\
\Leftrightarrow I = \dfrac12\cdot \dfrac{u^{2018}}{2018}\Bigg|_2^3\\
\Leftrightarrow I = \dfrac{3^{2018} - 2^{2018}}{4036}
\end{array}\)
