Найти первообразную функции

Решите задачу:

$4) \int (2x^5-5x)\, dx = \frac{2x^6}{6} - \frac{5x^2}{2} +C= \frac{x^6}{3} -2,5x^2+C \\5) \int (3cosx-x)\, dx=3sinx- \frac{x^2}{2} +C\\6)\int (cos(5x)- \frac{1}{6}sin(3x) )\, dx=\int (cos(5x))\,dx- \frac{1}{6} \int (sin(3x))\, dx \\3x=u \\3dx=du \\dx= \frac{du}{3} \\- \frac{1}{6}\int (sin(3x))\, dx= - \frac{1}{6}* \frac{1}{3} \int (sin(u))\, du= -\frac{1}{18} \int (sin(u))\, du=\\= \frac{cos(u)}{18} +C\\5x=t \\5dx=dt \\dx= \frac{dt}{5} \\\int (cos(5x))\,dx= \frac{1}{5}(\int (cos(t))\,dt)=\frac{sin(t)}{5}+C$
$\frac{sin(t)}{5}+\frac{cos(u)}{18}+C= \frac{sin(5x)}{5} + \frac{cos(3x)}{18} +C$