\right ]=\int_{1}^{3}{\mathrm{du}\over u}=\ln|u|_{1}^{3}=\ln|3|-\ln|1|=\ln3\\ \int_{0}^{\pi\over2}(\sin{x}-{2\over\pi}x)\mathrm{dx}=-\cos{x}|_{0}^{\pi\over2}-{x^2\over\pi}|_{0}^{\pi\over2}=1-{1\over\pi}({\pi^2\over4}-0)=1-{\pi\over4}\\ $$" alt="$$\Large \int_{0}^{\pi\over2}\cos{x}\mathrm{dx}=\sin{x}|_{0}^{\pi\over2}=1\\ \int_{0}^{1}\left ( x+{1\over\sqrt{x^2+1}} \right )\mathrm{dx}={1\over2}x^2|_{0}^{1}+\ln{\left (x+\sqrt{x^2+1} \right )}|_{0}^{1}={1\over2}+\ln{(1+\sqrt{1+1})}-\ln{(0+\sqrt{0+1})}={1\over2}+\ln{(1+\sqrt{2})}\\ \int_{0}^{1}{4x\over1+2x^2}\mathrm{dx}=\left [ 2x^2+1=u, dx={du\over4x};u\in<1;3> \right ]=\int_{1}^{3}{\mathrm{du}\over u}=\ln|u|_{1}^{3}=\ln|3|-\ln|1|=\ln3\\ \int_{0}^{\pi\over2}(\sin{x}-{2\over\pi}x)\mathrm{dx}=-\cos{x}|_{0}^{\pi\over2}-{x^2\over\pi}|_{0}^{\pi\over2}=1-{1\over\pi}({\pi^2\over4}-0)=1-{\pi\over4}\\ $$" align="absmiddle" class="latex-formula">
