Ответ:

[tex]1)\ \displaystyle \int \frac{cosx\, dx}{3sinx-1}=\Big[\ t=3sinx-1\ ,\ dt=3\, cosx\, dx\ \Big]=\frac{1}{3}\int \frac{dt}{t}=\\\\\\=\frac{1}{3}\cdot ln|t|+C=\frac{1}{3}\cdot ln|3sinx-1|+C\\\\\\2)\ \ \int\sqrt{(3z^4+2)^3}\cdot z^3\, dz=\Big[\ t=3z^4+2\ ,\ dt=12z^3\, dz\ \Big]=\frac{1}{12}\int \sqrt{t^3}\, dt=\\\\\\=\frac{1}{12}\int t^{3/2}\, dt=\frac{1}{12}\cdot \frac{t^{5/2}}{5/2}+C=\frac{1}{30}\cdot \sqrt{(3z^4+2)^5}+C[/tex]

[tex]3)\ \displaystyle \int \frac{x^2\, dx}{3x^3+1}=\Big[\ t=3x^3+1\ ,\ dt=9x^2\, dx\ \Big]=\frac{1}{9}\int \frac{dt}{t}=\frac{1}{9}\cdot ln|t|+C=\\\\\\=\frac{1}{9}\cdot ln|3x^3+1|+C[/tex]

P.S. Возможно условие такое:  

[tex]\displaystyle \int \frac{x^2\, dx}{3x^2+1}=\frac{1}{3} \int \Big(1-\frac{1}{3x^2+1}\Big)\, dx=\frac{1}{3}\int dx-\frac{1}{3\sqrt3}\int \frac{\sqrt3\, dx}{(\sqrt3x)^2+1}=\\\\\\=\frac{1}{3}\, x-\frac{1}{3\sqrt3}\cdot arctg(\sqrt3x)+C[/tex]