Chọn phương án A.

\(\begin{aligned}
\displaystyle\int\limits_1^3\dfrac{(x+6)^{2017}}{x^{2019}}\mathrm{\,d}x&=\displaystyle\int\limits_1^3 \dfrac{(x+6)^{2017}}{x^{2017}}\cdot\dfrac{1}{x^2}\mathrm{\,d}x\\
&=\displaystyle\int\limits_1^3 \left(\dfrac{x+6}{x}\right)^{2017}\cdot\dfrac{1}{x^2}\mathrm{\,d}x\\
&=\displaystyle\int\limits_1^3 \left(1+\dfrac{6}{x}\right)^{2017}\cdot\dfrac{1}{x^2}\mathrm{\,d}x.
\end{aligned}\)

Đặt \(u=1+\dfrac{6}{x}\) ta có

• \(\mathrm{d}u=-\dfrac{6}{x^2}\mathrm{\,d}x\Leftrightarrow \dfrac{1}{x^2}\mathrm{\,d}x=-\dfrac{1}{6}\mathrm{\,d}u\)
• \(x=1\Rightarrow u=7\)
• \(x=3\Rightarrow u=3\)

Khi đó $$\begin{aligned}
\displaystyle\int\limits_1^3\dfrac{(x+6)^{2017}}{x^{2019}}\mathrm{\,d}x&=-\dfrac{1}{6}\displaystyle\int\limits_7^3u^{2017}\mathrm{\,d}u\\
&=-\dfrac{u^{2018}}{6\cdot2018}\bigg|_7^3\\
&=\dfrac{7^{2018}-3^{2018}}{6\cdot2018}.
\end{aligned}$$
Suy ra \(a=7\).