Question

Помогите решить интеграл

Answer 1

$\int \frac{(\sqrt{3x+1}-1)dx}{\sqrt[3]{3x+1}+\sqrt{3x+1}}=[3x+1=t^6\; ,\; x=\frac{1}{3}(t^6-1)\; ,\; dx=2t^5\, dx]=\\\\=\int \frac{(t^3-1)\, dt}{t^2+t^3}=\int (1-\frac{t^2+1}{t^2(t+1)}\, dt=Q\\\\\frac{t^2+1}{t^2(t+1)}=\frac{A}{t^2}+\frac{B}{t}+\frac{C}{t+1}=\frac{A(t+1)+Bt(t+1)+Ct}{t^2(t+1)}\\\\t^2+1=At+A+Bt^2+Bt+Ct\\\\t=-1:\; \; \frac{2}{1}=-C\; ,\; \; C=-2\; ,\\\\t=0:\; \; \frac{1}{1}=A\; ,\; A=1\; , \\\\t^2\, |\; 1=B\\\\Q=\int \frac{dt}{t^2}+\int \frac{dt}{t}-2\int \frac{dt}{t+1}=-\frac{1}{t}+\ln|t|-2\, \ln|t+1|+C=$

$=-\frac{1}{\sqrt[6]{3x+1}}+\ln\sqrt[6]{3x+1}-2\, \ln|\sqrt[6]{3x+1}+1|+C\; .$