Question

Найдите неопределенные интегралы

Answer 1

$1); ; int frac{cos2x, dx}{cosx-sinx}=int frac{(cos^2x-sin^2x)dx}{cosx-sinx}=int frac{(cosx-sinx)(cosx+sinx)}{cosx-sinx}dx=\\=int (cosx+sinx)dx=sinx-cosx+C\\2); ; int frac{ln^4x}{x}dx=int ln^4xcdot frac{dx}{x}}=[frac{dx}{x}=d(lnx)]= frac{ln^5x}{5}+C\\3); ; int frac{dx}{sqrt{(1-x^2)arcsinx}}=int frac{1}{sqrt{arcsinx}}cdot frac{dx}{sqrt{1-x^2}} =[frac{dx}{sqrt{1-x^2}}=d(arcsinx)]=\\=int frac{d(arcsinx)}{sqrt{arcsinx}} =2sqrt{arcsinx}+C$

$4); ; int frac{x, dx}{ sqrt{x^2+4x+4} }=int frac{x, dx}{sqrt{(x+2)^2}}=int frac{x, dx}{x+2}=int (1-frac{2}{x+2} )=\\=x-2cdot ln|x+2|+C\\5); ; int frac{(2x-3)}{3x^2-7x+11}=int frac{(2x-3)dx}{3(x^2-frac{7}{3}x+frac{11}{3})}=frac{1}{3}int frac{(2x-3)dx}{(x-frac{7}{6})^2+frac{83}{36}} =\\=[t=x-frac{7}{6},; dt=dx,; x=t+frac{7}{6} ]=frac{1}{3}int frac{2t-frac{2}{3}}{t^2+frac{83}{36}}dt=\\=frac{1}{3}int frac{2t, dt}{t^2+frac{83}{36}}-frac{2}{9}int frac{dt}{t^2+frac{83}{36}}-frac{2}{9}int frac{dt}{t^2+frac{83}{36}}=$

$= frac{1}{3} cdot ln|t^2+ frac{83}{36}|-frac{2}{9}cdot frac{6}{sqrt{83}}cdot arctg frac{6t}{sqrt{83}}+C=\\= frac{1}{3} cdot ln|(x-frac{7}{6})^2+frac{83}{36}|-frac{4}{3sqrt{83}} cdot arctg frac{6(x-frac{7}{6})}{sqrt{83}}+C$