Question

Помогите решить интеграл

Answer 1

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

$=frac{1}{20}sqrt{(1+2x)^5} -frac{1}{16}sqrt{(1+2x)^4}-frac{1}{12}sqrt{(1+2x)^3}+frac{1}{8}sqrt{(1+2x)^2}+C$