Question

Найти следующие интегралы:

Answer 1

$\displaystyle\int(4x^2-1)^2dx=\int(16x^4-8x^2+1)dx=\frac{16x^5}{5}-\frac{8x^3}{3}+x+C\\\\\int\frac{3x+\sqrt{x}-2}{x}dx=\int(3+\frac{1}{\sqrt x}-\frac{2}{x})dx=3x+2\sqrt x-2ln|x|+C\\\\\int\frac{(x-2)(x-4)}{x^2}dx=\int\frac{x^2-6x+8}{x^2}dx\int(1-\frac{6}{x}+\frac{8}{x^2})dx=\\=x-6ln|x|-\frac{8}{x}+C$