Biết $\int \frac{{(x-1)}_{}^{2016}}{{(x+2)}_{}^{2018}}dx=\frac{1}{a}{\left(\frac{x-1}{x+2}\right)}_{}^b+C,x\ne -2$, với a, b nguyên dương. Mệnh đề nào dưới đây đúng?

A. a <  b

Ta có : . $\left(\frac{x-1}{x+2}\right)\text{'}=\frac{3}{{(x+2)}_{}^2}$

Khi đó:$\int \frac{{(x+1)}_{}^{2016}}{{(x+2)}_{}^{2018}}dx=\int {\left(\frac{x-1}{x+2}\right)}_{}^{2016}.\frac{1}{{(x+2)}_{}^2}dx$

Đặt $t=\frac{x-1}{x+2}\Rightarrow dt=\frac{3}{{(x+2)}_{}^2}dx\iff \frac{dx}{{(x+2)}_{}^2}=\frac{dt}{3}$$\begin{array}{l}\Rightarrow \int \frac{{(x+1)}_{}^{2016}}{{(x+2)}_{}^{2018}}dx=\int t_{}^{2016}\frac{dt}{3}=\frac{1}{3}\frac{t_{}^{2017}}{2017}+C=\frac{1}{6051}{\left(\frac{x-1}{x+2}\right)}_{}^{2017}+C\\ \Rightarrow a=\mathrm{6051,}b=2017\Rightarrow a=3b\end{array}$

Chọn: C