Question

Интеграл xdx/sqrt(2+3x-2x^2)
Интеграл ((x-4)dx)/(5x^2-x+7)

Answer 1

$1); ; int frac{dx}{sqrt{2+3x-2x^2}}=[, 2+3x-2x^2=-2(x^2-frac{3}{2}x-1)=\\=-2((x-frac{3}{4})^2-frac{9}{16}-1)=-2((x-frac{3}{4})^2-frac{25}{16})=2(frac{25}{16}-(x-frac{3}{4})^2, ]=\\=frac{1}{2}int frac{dx}{sqrt{frac{25}{16}-(x-frac{3}{4})^2}}=[, t=x-frac{3}{4},; dx=dt, ]=frac{1}{2}int frac{dt}{sqrt{frac{25}{16}-t^2}}=\\=frac{1}{2}cdot arcsinfrac{t}{5/4}+C=frac{1}{2}cdot arcsinfrac{4(x-frac{3}{4})}{5}+C=frac{1}{2}cdot arcsinfrac{4x-12}{5}+C$

$2); ; int frac{(x-4)dx}{5x^2-x+7}=[, 5(x^2-frac{1}{5}x+frac{7}{5})=5((x-frac{1}{10})^2-frac{1}{100}+frac{7}{5})=\\=5((x-frac{1}{10})^2-frac{139}{100}), ]=frac{1}{5}int frac{(x-4)dx}{(x-0,1)^2-1,39}=[, t=x-0,1; ;; dt=dx, ]=\\=frac{1}{5}int frac{(t-3,9)dt}{t^2-1,39}=frac{1}{10}int frac{2t, dt}{t^2-1,39}-frac{3,9}{5}int frac{dt}{t^2-1,39}=\\=0,1cdot ln|t^2-1,39|-frac{39}{50}cdot frac{10}{2sqrt{139}}cdot lnBig |frac{t-sqrt{1,39}}{t+sqrt{1,39}}Big |+C=\\=0,1cdot ln|x^2-0,2x+1,4|-frac{39}{2sqrt{139}}cdot lnBig |frac{10x-1-sqrt{139}}{10x-1+sqrt{139}}Big |+C$