1.найти первую производную у=arctg e ^2x+ln(корень((е^2х )+1)/(е^2х -1)) 2.Найти...

Решите задачу:

$1)\; \; y=arctge^{2x}+ln \sqrt{\frac{e^{2x}+1}{e^{2x}-1} }\\\\y'=\frac{1}{1+(e^{2x})^2}\cdot 2e^{2x}+\sqrt{\frac{e^{2x}-1}{e^{2x}+1}}\cdot \frac{1}{2}\cdot \sqrt{ \frac{e^{2x}-1}{e^{2x}+1}}\cdot \frac{2e^{2x}(e^{2x}-1)-2e^{2x}(e^{2x}+1)}{(e^{2x}-1)^2}=\\\\= \frac{2e^{2x}}{1+e^{4x}}-\frac{e^{2x}-1}{2(e^{2x}+1)}\cdot \frac{-4e^{2x}}{(e^{2x}-1)^2}= \frac{2e^{2x}}{1+e^{4x}}-\frac{2e^{2x}}{e^{4x}-1}= -\frac{4e^{2x}}{e^{8x}-1}$

$2)\; \; \int \frac{dx}{x(3lnx+5)} =\int \Big (\frac{1}{3lnx+5}\cdot \frac{dx}{x}\Big )=\Big [\; t=3lnx+5\; ,\; dt=\frac{3\cdot dx}{x}\; \Big ]=\\\\=\frac{1}{3}\cdot \int \frac{dt}{t}= \frac{1}{3}\cdot ln|t|+C= \frac{1}{3}\cdot ln|3lnx+5|+C;\\\\\\(\frac{1}{3}\cdot ln|3lnx+5|+C)'= \frac{1}{3}\cdot \frac{1}{3lnx+5}\cdot 3\cdot \frac{1}{x}+0= \frac{1}{x(3lnx+5)} \; .$