$7)\\ A=\int\limits^{+\infty}_0 xe^{-2x} \, dx\\ = \displaystyle\lim_{t \to +\infty} \displaystyle\int\limits^{t}_0 xe^{-2x} \, dx\\ = \displaystyle\lim_{t \to +\infty} I\\ I=\displaystyle\int\limits^{t}_0 xe^{-2x} \, dx\\ u=x \Rightarrow du=dx\\ dv=e^{-2x} \, dx \Rightarrow v=-\dfrac{1}{2}e^{-2x}\\ I=-\dfrac{1}{2}xe^{-2x}\Bigg\vert^{t}_0+\dfrac{1}{2}\displaystyle\int\limits^{t}_0 e^{-2x} \, dx\\ =-\dfrac{1}{2}xe^{-2x}\Bigg\vert^{t}_0-\dfrac{1}{4}\displaystyle\int\limits^{t}_0 e^{-2x} \, d(-2x)\\ =-\dfrac{1}{2}te^{-2t}-\dfrac{1}{4} e^{-2x} \Bigg\vert^{t}_0\\ =-\dfrac{1}{2}te^{-2t}-\dfrac{1}{4} e^{-2t}+\dfrac{1}{4}\\ =-\dfrac{1}{2}e^{-2t}\left(t+\dfrac{1}{2}\right)+\dfrac{1}{4}\\ A= \displaystyle\lim_{t \to +\infty} \left(-\dfrac{1}{2}e^{-2t}\left(t+\dfrac{1}{2}\right)+\dfrac{1}{4}\right)\\ =\dfrac{1}{4}\\ 8)\\ B=\displaystyle\int\limits^{+\infty}_1 \dfrac{\ln x}{x^3} \, dx\\ =\displaystyle\lim_{t \to +\infty} \displaystyle\int\limits^{t}_1 \dfrac{\ln x}{x^3} \, dx\\ =\displaystyle\lim_{t \to +\infty} I\\ I=\displaystyle\int\limits^{t}_1 \dfrac{\ln x}{x^3} \, dx\\ u=\ln x \Rightarrow du=\dfrac{1}{x}dx\\ dv=\dfrac{dx}{x^3} \Rightarrow v=\dfrac{-1}{2x^2}\\ I=-\dfrac{\ln x}{2x^2} \Bigg\vert^{t}_1 +\dfrac{1}{2}\displaystyle\int\limits^{t}_1 \dfrac{1}{x^3} \, dx\\ =-\dfrac{\ln t}{2t^2}-\dfrac{1}{4x^2}\Bigg\vert^{t}_1\\ =-\dfrac{\ln t}{2t^2}-\dfrac{1}{4t^2}+\dfrac{1}{4}\\ B=\displaystyle\lim_{t \to +\infty} \left(-\dfrac{\ln t}{2t^2}-\dfrac{1}{4t^2}+\dfrac{1}{4}\right)\\ =\dfrac{1}{4}\\ \left(\displaystyle\lim_{t \to +\infty} -\dfrac{\ln t}{2t^2}=\displaystyle\lim_{t \to +\infty} -\dfrac{(\ln t)'}{(2t^2)'}=\displaystyle\lim_{t \to +\infty} -\dfrac{1}{4t^2}=0(L'Hospital)\right)$
