Решите пожалуйста (те кто разбирается!!!)

Решите задачу:

$y=x\cos(3x^3), \\ dy=(x\cos(3x^3))'dx=(x'\cos(3x^3)+x(\cos(3x^3))')dx= \\ =(\cos(3x^3)-x\sin(3x^3)\cdot(3x^3)')dx= \\ =(\cos(3x^3)-x\sin(3x^3)\cdot(9x^2))dx=(\cos(3x^3)-9x^3\sin(3x^3))dx;$

$y= \frac{\sin^4x}{\sqrt{2x}}, \\ dy = (\frac{\sin^4x}{\sqrt{2x}})'dx = \frac{(\sin^4x)'\sqrt{2x}-\sin^4x(\sqrt{2x})'}{(\sqrt{2x})^2}dx = \\ = \frac{4\sin^3x(\sin x)'\sqrt{2x}-\sin^4x\cdot \frac{\sqrt{2}}{2\sqrt{x}}}{2x}dx = \frac{4\sin^3x\cos x\sqrt{2x}-\sin^4x\cdot \frac{1}{\sqrt{2x}}}{2x}dx.$

$\int {\sin(3x^3+6x^2)(3x^2+4x)} \, dx = \frac{1}{3} \int {\sin(3x^3+6x^2)} \, d(3x^3+6x^2) =\\= - \frac{1}{3}\cos(3x^3+6x^2).$

$\int\limits^1_0 {e^{x^2+1}} \, dx = \frac{1}{2} \int\limits^1_0 {e^{x^2+1}} \, d(x^2+1) = \frac{1}{2}e^{x^2+1}|^1_0= \frac{1}{2}e^{2}+ \frac{1}{2}e= \frac{1}{2}e(e+1);$