Question

Помогите пожалуйста!!​

Answer 1

Ответ:

решение на фотографии.

Answer 2

Ответ:

$u'x=\dfrac{5-4u^2}{3+4u}\ \ ,\ \ \ \ \dfrac{du}{dx}\cdot x=\dfrac{5-4u^2}{3+4u}\ \ ,\ \ \ \int \dfrac{(3+4u)\, du}{5-4u^2}=\int \dfrac{dx}{x}\ .\\\\\\\int \dfrac{(3+4u)\, du}{5-4u^2}=\int \dfrac{3\, du}{5-4u^2}+\int \dfrac{4u\, du}{5-4u^2}=\dfrac{1}{2}\int \dfrac{3\cdot 2du}{5-(2u)^2}-\dfrac{4}{8}\int \dfrac{-8\, du}{5-4u^2}=\\\\\\=\dfrac{3}{2}\cdot \dfrac{1}{2\sqrt5}\cdot ln\Big|\, \dfrac{\sqrt5+2u}{\sqrt5-2u}\, \Big|-\dfrac{1}{2}\cdot ln\Big|\, 5-4u^2\, \Big|+lnC_1\ ;$

$\dfrac{3}{4\sqrt5}\cdot ln\Big|\, \dfrac{\sqrt5+2u}{\sqrt5-2u}\, \Big|-\dfrac{1}{2}\cdot ln\Big|\, 5-4u^2\, \Big|=ln|x|+lnC\\\\\\\dfrac{3}{4\sqrt5}\cdot ln\Big|\, \dfrac{\sqrt5+2u}{\sqrt5-2u}\, \Big|-\dfrac{1}{2}\cdot ln\Big|\, 5-4u^2\, \Big|+C=ln|\, Cx\, |$