Question

14 Вариант,11 и 15 задания

Answer 1

$11); ; int frac{arcsin^22x}{sqrt{1-4x^2}}, dx=[, t=arcsin2x,; dt=frac{2, dx}{sqrt{1-4x^2}}]=\\=frac{1}{2}int t^2, dt=frac{1}{2}cdot frac{t^3}{3}+C=frac{1}{6}cdot arctg^32x+C\\15); ;int frac{dx}{3x^2+5x+1}=int frac{dx}{3(x^2+frac{5}{3}cdot x+frac{1}{3})}=frac{1}{3}int frac{dx}{(x+frac{5}{6})^2-frac{13}{36}}=\\=[, t=x+frac{5}{6},; dt=dx, ]=frac{1}{3}int frac{dt}{t^2-frac{13}{36}}=frac{1}{3}cdot frac{6}{2sqrt{13}}cdot lnBig |frac{t-frac{sqrt{13}}{6}}{t+frac{sqrt{13}}{6}}Big |+C=$

$=frac{1}{sqrt{13}}cdot lnBig |frac{x+frac{5}{6}-frac{sqrt{13}}{6}}{x+frac{5}{6}+frac{sqrt{13}}{6}}Big |+C=frac{1}{sqrt{13}}cdot lnBig |frac{6x+5-sqrt{13}}{6x+5+sqrt{13}}Big |+C$