$\int_1^2 \dfrac{1}{x(x^{10}+1)^2}dx=\int_1^2 \dfrac{x^9}{x^{10}(x^{10}+1)^2}dx$
đặt $x^{10}+1 = t \Rightarrow 10x^9 dx = dt$
$I = \dfrac{1}{10}\int \dfrac{1}{(t-1).t^2}dt =\dfrac{1}{10} \int \bigg (-\dfrac{1}{t^2}-\dfrac{1}{t}+\dfrac{1}{t-1} \bigg )dt$
Toàn tp dễ rồi tự làm