Đặt $x=tanu$$\Rightarrow \begin{cases}dx=\frac{du}{cos^2u}\\ x^2\sqrt{x^2+1}=tan^2u.\sqrt{tan^2u+1}=\frac{tan^2u}{cosu} \end{cases}$Đổi cận nhé$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{tan^2u.cos^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{du}{tan^2u.cosu}$$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{sin^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{d(sinu)}{sin^2u}$$\Rightarrow -\int\limits_{\frac{\pi }{4}}^{arctan2}d(\frac{1}{sinu}) =\int\limits_{arctan2}^{\frac{\pi }{4}}d(\frac{1}{sinu})$
Đặt $x=tanu$$\Rightarrow \begin{cases}dx=\frac{du}{cos^2u}\\ x^2\sqrt{x^2+1}=tan^2u.\sqrt{tan^2u+1}=\frac{tan^2u}{cosu} \end{cases}$Đổi cận nhé$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{tan^2u.cos^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{du}{tan^2u.cosu}$$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{sin^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{d(sinu)}{sin^2u}$$\Rightarrow -\int\limits_{\frac{\pi }{4}}^{arctan2}d(\frac{1}{sinu}) $
Đặt $x=tanu$$\Rightarrow \begin{cases}dx=\frac{du}{cos^2u}\\ x^2\sqrt{x^2+1}=tan^2u.\sqrt{tan^2u+1}=\frac{tan^2u}{cosu} \end{cases}$Đổi cận nhé$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{tan^2u.cos^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{du}{tan^2u.cosu}$$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{sin^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{d(sinu)}{sin^2u}$$\Rightarrow -\int\limits_{\frac{\pi }{4}}^{arctan2}d(\frac{1}{sinu})
=\int\limits_{arctan2}^{\frac{\pi }{4}}d(\frac{1}{sinu})$