Question

Решить неопределенный интеграл

Answer 1

Ответ:

$\dfrac{4}{3}\sqrt[4]{x^{3}}+\dfrac{1}{2}ln \dfrac{1+\sqrt[8]{x}}{1-\sqrt[8]{x}}+C, \quad C-const;$

Пошаговое объяснение:

$\int\ {\dfrac{dx}{\sqrt[4]{x}-\sqrt{x}}} \, = \int\ {\dfrac{1}{x^{\tfrac{1}{4}}-x^{\tfrac{1}{2}}}} \, dx = \int\ {\dfrac{1}{x^{\tfrac{1}{4}}-x^{\tfrac{2}{4}}}} \, dx = \int\ {\dfrac{1}{x^{\tfrac{1}{4}}(1-x^{\tfrac{1}{4}})}} \, dx ;$

$\dfrac{1}{x^{\tfrac{1}{4}}(1-x^{\tfrac{1}{4}})}=\dfrac{A}{x^{\tfrac{1}{4}}}+\dfrac{B}{1-x^{\tfrac{1}{4}}};$

$\dfrac{1}{x^{\tfrac{1}{4}}(1-x^{\tfrac{1}{4}})}=\dfrac{A(1-x^{\tfrac{1}{4}})+Bx^{\tfrac{1}{4}}}{x^{\tfrac{1}{4}}(1-x^{\tfrac{1}{4}})};$

$A(1-x^{\tfrac{1}{4}})+Bx^{\tfrac{1}{4}}=1;$

$A-Ax^{\tfrac{1}{4}}+Bx^{\tfrac{1}{4}}=1;$

$A+Bx^{\tfrac{1}{4}}-Ax^{\tfrac{1}{4}}=1;$

$A+(B-A)x^{\tfrac{1}{4}}=1+0x^{\tfrac{1}{4}};$

$\displaystyle \left \{ {{A=1} \atop {B-A=0}} \right. \Leftrightarrow \left \{ {{A=1} \atop {B=A}} \right. \Leftrightarrow \left \{ {{A=1} \atop {B=1}} \right. ;$

$\dfrac{1}{x^{\tfrac{1}{4}}(1-x^{\tfrac{1}{4}})}=\dfrac{1}{x^{\tfrac{1}{4}}}+\dfrac{1}{1-x^{\tfrac{1}{4}}}=x^{-\tfrac{1}{4}}+\dfrac{1}{1-(x^{\tfrac{1}{8}})^{2}};$

$\int\ {\dfrac{1}{x^{\tfrac{1}{4}}(1-x^{\tfrac{1}{4}})}} \, dx = \int\ {\Bigg (x^{-\tfrac{1}{4}}+\dfrac{1}{1-(x^{\tfrac{1}{8}})^{2}}} \Bigg ) \, dx = \int\ {x^{-\tfrac{1}{4}}} \, dx + \int\ {\dfrac{1}{1-(x^{\tfrac{1}{8}})^{2}}} \, dx =$

$=\dfrac{x^{-\tfrac{1}{4}+1}}{-\dfrac{1}{4}+1}+arth (x^{\tfrac{1}{8}})+C=\dfrac{x^{\tfrac{3}{4}}}{\dfrac{3}{4}}+\dfrac{1}{2}ln \dfrac{1+x^{\tfrac{1}{8}}}{1-x^{\tfrac{1}{8}}}+C=\dfrac{4}{3}x^{\tfrac{3}{4}}+\dfrac{1}{2}ln \dfrac{1+x^{\tfrac{1}{8}}}{1-x^{\tfrac{1}{8}}}+C=$

$=\dfrac{4}{3}\sqrt[4]{x^{3}}+\dfrac{1}{2}ln \dfrac{1+\sqrt[8]{x}}{1-\sqrt[8]{x}}+C, \quad C-const;$