C1:tích phân từ 1 đến 2 của: dx/(x^2.căn(1+x^2)
$\int_{1}^{2}\dfrac{dx}{x^2\sqrt{1+x^2}}$
Đặt $xt = \sqrt{1+x^2} \rightarrow t = \dfrac{\sqrt{x^2+1}}{x}$
$x = 2 \rightarrow t = \dfrac{\sqrt{5}}{2}$
$x = 1 \rightarrow t = \sqrt{2}$
$(xt)^2 = x^2+1 \rightarrow (t^2-1)x^2 = 1 \rightarrow x^2 = \dfrac{1}{t^2-1} \rightarrow xdx = \dfrac{-tdt}{(t^2-1)^2}$
$\dfrac{dx}{\sqrt{x^2+1}} = \dfrac{x dx}{x.\sqrt{x^2+1}} = \dfrac{ x dx}{x.tx} = \dfrac{\dfrac{-t dt}{(t^2+1)^2}}{\dfrac{t}{t^2+1}} = \dfrac{-dt}{t^2-1}$
$C = \int_{\sqrt{2}}^{\dfrac{\sqrt{5}}{2}}\dfrac{-dt}{\dfrac{1}{t^2-1}(t^2-1)}$