Tính tích phân 

B

- Hướng dẫn giải

Đáp án B

Ta có 

$\begin{array}{l}I=\underset{1}{\overset{2}{\int }}\frac{{\left(x+2\right)}^{2017}}{x^{2019}}dx\\ =\underset{1}{\overset{2}{\int }}{\left(1+\frac{2}{x}\right)}^{2017}.\frac{1}{x^2}dx\end{array}$

Đặt 

$\begin{array}{l}t=1+\frac{2}{x}\\ \Rightarrow \left\{\begin{array}{l}x=\frac{2}{t-1}\Rightarrow dx=-\frac{2}{{\left(t-1\right)}^2}dt\\ x^2=\frac{4}{{\left(t-1\right)}^2}\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}x=1\Rightarrow t=3\\ x=2\Rightarrow t=2\end{array}\right.\end{array}$

Suy ra

$\begin{array}{l}I=-\underset{3}{\overset{2}{\int }}\frac{t^{2017}.2{\left(t-1\right)}^2}{4{\left(t-1\right)}^2}dt\\ =\frac{1}{2}\underset{2}{\overset{3}{\int }}{t}^{2017}dt={\left.\frac{t^{2018}}{4036}\right|}_2^3\\ =\frac{3^{2018}-2^{2018}}{4036}\end{array}$