Ответ:
Найти интеграл . Применяем метод замены .
[tex]\bf \displaystyle \int \frac{e^{2x}\, dx}{\sqrt[4]{\bf e^{x}+1}}=\int \frac{e^{x}\cdot e^{x}\, dx}{\sqrt[4]{\bf e^{x}+1}}=\\\\\\=\Big[\ t^4=e^{x}+1\ \ ,\ \ e^{x}=t^4-1\ ,\ e^{x}\, dx=4t^3\, dt\ ,\ t=\sqrt[4]{\bf e^{x}+1}\ \Big]=\\\\\\=\int \frac{(t^4-1)\cdot 4t^3\, dt}{\sqrt[4]{\bf t^4}}=4\int \frac{(t^7-t^3)\, dt}{t}=4\int (t^6-t^2)\, dt=\\\\\\=4\, \Big(\frac{t^7}{7}-\frac{t^3}{3}\Big)+C=\frac{4}{7}\, \sqrt[4]{\bf (e^{x}+1)^7}-\frac{4}{3}\, \sqrt[4]{\bf (e^{x}+1)^3}+C[/tex]
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Ответ:
Найти интеграл . Применяем метод замены .
[tex]\bf \displaystyle \int \frac{e^{2x}\, dx}{\sqrt[4]{\bf e^{x}+1}}=\int \frac{e^{x}\cdot e^{x}\, dx}{\sqrt[4]{\bf e^{x}+1}}=\\\\\\=\Big[\ t^4=e^{x}+1\ \ ,\ \ e^{x}=t^4-1\ ,\ e^{x}\, dx=4t^3\, dt\ ,\ t=\sqrt[4]{\bf e^{x}+1}\ \Big]=\\\\\\=\int \frac{(t^4-1)\cdot 4t^3\, dt}{\sqrt[4]{\bf t^4}}=4\int \frac{(t^7-t^3)\, dt}{t}=4\int (t^6-t^2)\, dt=\\\\\\=4\, \Big(\frac{t^7}{7}-\frac{t^3}{3}\Big)+C=\frac{4}{7}\, \sqrt[4]{\bf (e^{x}+1)^7}-\frac{4}{3}\, \sqrt[4]{\bf (e^{x}+1)^3}+C[/tex]