$\begin{array}{l}1)\quad \displaystyle\int\dfrac{e^x}{1 + e^{2x}}dx\\ Đặt\,\,u = e^x\\ \to du = e^xdx\\ \text{Ta được:}\\ \displaystyle\int\dfrac{du}{1+u^2}\\ = \arctan u + C\\ = \arctan(e^x) + C\\ 2)\quad \displaystyle\int\dfrac{dx}{x\sqrt{x^2+1}}\\ Đặt\,\,u = x^2 +1\\ \to du = 2xdx\\ \text{Ta được:}\\ \displaystyle\int\dfrac{du}{2(u-1)\sqrt u}\\ Đặt\,\,t = \sqrt u\\ \to dt = \dfrac{1}{2\sqrt u}du\\ \text{Ta được:}\\ -\displaystyle\int\dfrac{dt}{1- t^2}\\ = -\dfrac{\ln|1+t| - \ln|1-t|}{2}\\ = \dfrac12\ln\left|\dfrac{1-t}{1+t} \right|\\ 3)\quad \displaystyle\int\dfrac{\sqrt{1+\ln^2x}.\ln x}{x}dx\\ Đặt\,\,u = \ln x\\ \to du = \dfrac{dx}{x}\\ \text{Ta được:}\\ \displaystyle\int u\sqrt{u^2+1}du\\ Đặt\,\,t = u^2 +1\\ \to dt = 2udu\\ \text{Ta được:}\\ \dfrac12\displaystyle\int\sqrt tdt\\ = \dfrac12\cdot\dfrac{2\sqrt{t^3}}{3}+C\\ = \dfrac{\sqrt{(u^2+1)^3}}{3} +C\\ = \dfrac{\sqrt{(\ln^2x+1)^3}}{3} +C\\ \end{array}$
