Question

Problema 4 și 7 vă rog mult!!! ​

Answer 1

$4)\quad \displaystyle \int \dfrac{1}{x(x^{2015}+1)}\, dx =\int \dfrac{x^{2014}}{x^{2015}(x^{2015}+1)}\, dx\overset{(*)}{=}\\ \\\\ x^{2015} = t \Rightarrow 2015x^{2014}\, dx = dt \Rightarrow x^{2014}\, dx = \dfrac{1}{2015}\, dt\\ \\ \\ \overset{(*)}{=}\int \dfrac{1}{t(t+1)}\cdot \dfrac{1}{2015}\, dt = \dfrac{1}{2015}\Big(\int \dfrac{1}{t}\, dt - \int \dfrac{1}{t+1}\, dt\Big) = \\ \\ = \dfrac{1}{2015}\Big(\ln t-\ln (t+1)\Big)+C = \dfrac{1}{2015}\ln \dfrac{t}{t+1}+C = \\ \\ = \dfrac{1}{2015}\ln \dfrac{x^{2015}}{1+x^{2015}}+C$

$7)\quad |z-a| = \sqrt{a^2-b^2} \\ \\ \sqrt{(z-a)(\overline{z-a})} = \sqrt{a^2-b^2} \\ (z-a)(\overline{z-a}) = a^2-b^2 \\ (z-a)(\overline{z}-a) = a^2-b^2\\ z\overline{z}-az-a\overline{z}+a^2 = a^2-b^2 \\\\ \Rightarrow z\overline{z} = az+a\overline{z}-b^2 \\ \\ \Big|\dfrac{b-z}{b+z}\Big| = \dfrac{|b-z|}{|b+z|} = \sqrt{\dfrac{(b-z)(\overline{b-z})}{(b+z)(\overline{b+z})}} = \sqrt{\dfrac{(b-z)(b-\overline{z})}{(b+z)(b+\overline{z})}} =$

$= \sqrt{\dfrac{b^2-b\overline{z}-bz+z\overline{z}}{b^2+b\overline{z}+bz+z\overline{z}}} = \sqrt{\dfrac{b^2-b\overline{z}-bz+az+a\overline{z}-b^2}{b^2+b\overline{z}+bz+az+a\overline{z}-b^2}}= \\ \\ = \sqrt{\dfrac{\overline{z}(a-b)+z(a-b)}{\overline{z}(a+b)+z(a+b)}} = \sqrt{\dfrac{(a-b)(\overline{z}+z)}{(a+b)(\overline{z}+z)}} = \sqrt{\dfrac{a-b}{a+b}}$