Помогите пожалуйста проинтегрировать уравнения?

Решите задачу:

$\LARGE \int \frac{(4-x)^2}{4\sqrt{x}}\mathrm{d}x = \int \frac{16-8x+x^2}{4\sqrt{x}}\mathrm{d}x = \int \frac{16}{4\sqrt{x}}\mathrm{d}x+\int \frac{8x}{4\sqrt{x}}\mathrm{d}x+\int \frac{x^2}{4\sqrt{x}}\mathrm{d}x = 4\int \frac{\mathrm{d}x}{\sqrt{x}} +2\int \frac{x}{\sqrt{x}}\mathrm{d}x+\frac{1}{4}\int \frac{x^2}{\sqrt{x}}\mathrm{d}x=8\sqrt{x}+\frac{4x^{\frac{3}{2}}}{3}+\frac{x^{\frac{5}{2}}}{10}+C\\$

$\LARGE \int \frac{5*3^x-4*2^x}{3^x}\mathrm{d}x = 5\int \frac{3^x}{3^x}\mathrm{d}x - 4\int \frac{2^x}{3^x}\mathrm{d}x = 5\int \mathrm{d}x - 4\int (\frac{2}{3})^{x}\mathrm{d}x = 5x-\frac{(\frac{2}{3})^{x}}{\ln{\frac{2}{3}}}+C$

$\LARGE \int \frac{\mathrm{d}x}{\sqrt{2-2x^2}}=\frac{1}{\sqrt2}\arcsin\frac{\sqrt{2x}}{\sqrt2}+C=\frac{1}{\sqrt2}\arcsin\sqrt{x}+C$