Question

ОЧЕЕЕЕЕНЬ СРОЧНО, ПОМОГИТЕ ПАЗЯЗЯ!!
1)int x^2dx/x^4+5x^2 +4

2)int x arctg2xdx

3)int dx/(x+1) sqrt 1-x-x^2

4)int sin^3 5xdx

5)int 5sqrtx -2x^3 +4/x^2 .dx

6)int 2x+3/sqrt 2x^2-x+6

Answer 1

$2); ; int xcdot arctg2x, dx=[, u=arctg2x,; du=frac{2, dx}{1+4x^2},; dv=x, dx,; v=frac{x^2}{2}, ]=\\=frac{x^2}{2}cdot arctg2x-int frac{x^2, dx}{1+4x^2}=frac{x^2}{2}cdot arctg2x-frac{1}{4}int frac{x^2, dx}{x^2+frac{1}{4}}=\\=frac{x^2}{2}cdot arctg2x-frac{1}{4}int (1-frac{1}{4}cdot frac{1}{x^2+frac{1}{4}})dx=\\=frac{x^2}{2}cdot arctg2x-frac{1}{4}x+frac{1}{16}cdot 2, arctg, 2x+C; ;$

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

$6); ; int frac{(2x+3)dx}{sqrt{2x^2-x+6}}=int frac{(2x+3)dx}{sqrt{2cdot (, (x-frac{1}{4})^2+frac{47}{16})}}=[, t=x-frac{1}{4},; x=t+frac{1}{4},; dx=dt, ]=\\=frac{1}{sqrt2}int frac{(2t+3,5)dt}{sqrt{t^2+frac{47}{16}}}=frac{1}{sqrt2}int frac{2t, dt}{sqrt{t^2+frac{47}{16}}}+frac{3,5}{sqrt2}int frac{dt}{sqrt{t^2+frac{47}{16}}}=\\=frac{1}{sqrt2}cdot 2, sqrt{t^2+frac{47}{16}}+frac{7}{2sqrt2}cdot lnBig |t+sqrt{t^2+frac{47}{16}}Big |+C=$

$=sqrt2cdot sqrt{x^2-frac{x}{2}+3}+frac{7}{2sqrt2}cdot lnBig |x-frac{1}{4}+sqrt{x^2-frac{x}{2}+3}, Big |+C; .$