Решите срочно!!!!!!!

Решите задачу:

$1)\; \int \sqrt[4]{(3+5x^4)^3}\cdot x^3dx=[t=3+5x^4,\; dt=20x^3dx]=\int t^{\frac{3}{4}}\cdot \frac{1}{20}dt=\\\\=\frac{1}{20}\frac{t^{\frac{7}{4}}}{\frac{7}{4}}+C=\frac{1}{35}(3+5x)^{\frac{7}{4}}+C\\\\2)\; \int (2x^3-1)^2\cdot x^2dx=[t=2x^3-1,\; dt=6x^2dx]=\int t^2\cdot \frac{1}{6}dt=\\\\=\frac{1}{6}\cdot \frac{t^3}{3}+C=\frac{1}{18}\cdot (2x^3-1)+C$

$3)\; \int \frac{x+3}{x^2-4x+3}dx=\int \frac{x+3}{(x-2)^2-1}dx=[t=x-2,dx=dt]=\int \frac{t+5}{t^2-1}dt=\\\\=\frac{1}{2}\int \frac{2t\cdot dt}{t^2-1}+\int \frac{5}{t^2-1}dt=\frac{1}{2}ln|t^2-1|+5\cdot \frac{1}{2}ln|\frac{t-1}{t+1}|+C=\\\\=\frac{1}{2}ln|x^2-4x+3| +\frac{5}{2}ln|\frac{x-3}{x-1}|+C$

$4)\; \int \frac{x^2+x+1}{x^2-5x+6}dx=\int (1+\frac{6x-5}{x^2-5x+6})dx=\int dx+\int \frac{6x-5}{(x-2)(x-3)}dx=\\\\\frac{6x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}=\frac{-7}{x-2}+\frac{13}{x-3}\\\\=x-7\int \frac{dx}{x-2}+13\int \frac{dx}{x-3}=x-7ln|x-2|+13ln|x-3|+C$