Tính tích phân $I= \int_1^2 \frac{2}{x^3+2x} dx$

Tính tích phân $I=\displaystyle \int_1^2 \dfrac{2}{x^3+2x} dx$


$I=\displaystyle \int_1^2 \dfrac{2}{x(x^2+2)} dx=\displaystyle \int_1^2 \dfrac{2x}{x^2(x^2+2)} dx$

Đặt $ t=x^2 \Rightarrow dt=2x.dx$

$x=1 \Rightarrow t=1 , \,\,  x=2 \Rightarrow t=4$

$I= \displaystyle \int_1^4 \dfrac{1}{t(t+2)}dt =\displaystyle \int_1^4 \left[ \dfrac{1}{2t}-\dfrac{1}{2(t+2)} \right] dt= \dfrac{1}{2} \ln \bigg| \dfrac{t}{t+2} \bigg| \bigg|_{1}^{4}=\dfrac{1}{2} \ln \dfrac{4}{3}$