Verified answer

Ответ:

13)  Интегрирование иррациональных выражений . Метод замены .

[tex]\bf \displaystyle \int \frac{2x+\sqrt{x+1}}{1+\sqrt{x+1}}\, dx=\Big[\ t=\sqrt{x+1}\ ,\ t^2=x+1\ ,\ x=t^2-1\ ,\ dx=2t\, dt\ \Big]=\\\\\\=\int \frac{2(t^2-1)+t}{1+t}\cdot 2t\, dt=2\int \frac{2t^3+t^2-2t}{1+t}\, dt=\\\\\\=2\int \Big(2t^2-t-1+\frac{1}{1+t}\Big)\, dt=2\Big(\frac{2t^3}{3}-\frac{t^2}{2}-t+ln|\, 1+t\, |\Big)+C=\\\\\\=\frac{4\sqrt{(x+1)^3}}{3}-x-1-2\sqrt{x+1}+2ln\Big|\, 1+\sqrt{x+1}\,\Big|+C[/tex]

14)

[tex]\bf \displaystyle \int \frac{\sqrt[6]{\bf x}}{1+\sqrt[3]{\bf x}}\, dx=\Big[\ x=t^6\ ,\ dx=6\, t^5\, dt\ ,\ t=\sqrt[6]{\bf x}\ \Big]=\int \frac{t\cdot 6t^5}{1+t^2}\, dt=\\\\\\=6\int \frac{t^6}{t^2+1}\, dt=6\int \Big(t^4-t^2+1-\frac{1}{t^2+1}\Big)\, dt=\\\\\\=6\, \Big(\frac{t^5}{5}-\frac{t^3}{3}+t-arctg\, t\Big)+C=\frac{6\sqrt[6]{\bf x^5}}{5}-2\sqrt[6]{\bf x^3}+6\sqrt[6]{\bf x}-6\, arctg\sqrt[6]{\bf x}+C[/tex]