Решите пожалуйста, буду очень благодарен!

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

$\int {\frac{(1 - \ln(x))^4}{x}} \, dx, t = 1 - \ln(x), dt = \frac{dx}{x}\\\int {\frac{(1 - \ln(x))^4}{x}} \, dx = \int {t^4}\, dt = \frac{t^5}{5} + C = \frac{(1 - \ln (x))^5}{5} + C\\$

$\int {(5x + 4)^3} \, dx, t = 5x + 4, dt = 5dx\\\int {(5x + 4)^3} \, dx = \frac{1}{5}\int {t^3} \, dt = \frac{t^4}{20} + C = \frac{(5x + 4)^4}{20} + C$

$\int {\frac{x}{x^2 + 1}} \, dx, t = x^2 + 1, dt = 2xdx\\\int {\frac{x}{x^2 + 1}} \, dx = \frac{1}{2} \int {\frac{1}{t}} \, dt = \frac{\ln(t)}{2} + C = \frac{\ln(x^2 + 1)}{2} + C$

$\int {e^{\cos(x)}\sin(x)} \, dx, t = \cos(x), dt = -\sin(x)dx\\\int {e^{\cos(x)}\sin(x)} \, dx = -\int {e^t} dt = -e^t + C$