Question

14 Вариант,11 и 15 задания

Answer 1

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

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