Question

Решите кто понимает очень хочу понять как решается подобное.

Answer 1

$1); ; int frac{(6x-5), dx}{2sqrt{3x^2-5x+6}}=Big [; 3x^2-5x+6=3, (x^2-frac{5}{3}x+2)=3, Big ((x-frac{5}{6})^2-frac{25}{36}+2Big )=\\=3, Big ((x-frac{5}{6})^2- frac{47}{36}Big ); Big]=frac{1}{2sqrt3}int frac{(6x-5), dx}{sqrt{(x-frac{5}{6})^2-frac{47}{36}}}=Big[; t=x-frac{5}{6}; ,; dt=dx,$

$x=t+frac{5}{6}; Big]=frac{1}{2sqrt3}int frac{6t, dt}{sqrt{t^2-frac{47}{36}}}=frac{3}{2sqrt3}int frac{2, t, dt}{sqrt{t^2-frac{47}{36}}}=Big [; z=t^2-frac{47}{36}; ,; dz=2t, dt; Big]=\\=frac{sqrt3}{2}int frac{dz}{sqrt{z}}=frac{sqrt3}{2}cdot 2sqrt{z}+C=sqrt3cdot sqrt{x^2-frac{5}{3}x+2}+C; ;$

$2); ; int frac{sqrt{1-x^2}}{x^2}, dx=Big[; x=sint; ,; dx=cost, dt; ,; t=arcsinx; ,\\sqrt{1-x^2}=sqrt{1-sin^2t}=sqrt{cos^2t}=cost; Big]=int frac{cost}{sin^2t}cdot cost, dt=\\=int frac{cos^2t}{sin^2t}, dt=int frac{1-sin^2t}{sin^2t}, dt=int frac{dt}{sin^2t}-int dt=-ctgt-t+C=\\=-ctg(arcsinx)-arcsinx+C=frac{sqrt{1-x^2}}{x}-arcsinx+C$

$3); ; int frac{cos^3x}{sin^2x+sinx}, dx=int frac{cos^2xcdot cosx, dx}{sin^2x(sinx+1)}=int frac{(1-sin^2x)cdot cosx, dx}{sinx(sinx+1)}=\\Big[; t=sinx; ,; dt=cosx, dx; Big]=int frac{(1-t^2)cdot dt}{t(t+1)}=int frac{(1-t)(1+t)cdot dt}{tcdot (t+1)}=int frac{1-t}{t}, dt=\\=int (frac{1}{t}-1), dt=int frac{dt}{t}-int dt=ln|t|-t+C=ln|sinx|-sinx+C; ;$

$4); ; int frac{dx}{x(x^2+1)}=Big[; x=frac{1}{t}; ,; dx=-frac{dt}{t^2}; Big]=-int frac{frac{dt}{t^2}}{frac{1}{t}cdot (frac{1}{t^2}+1)}=-int frac{dt}{tcdot frac{1+t^2}{t^2}}=\\=-int frac{tcdot dt}{1+t^2}=-frac{1}{2}int frac{2tcdot dt}{1+t^2}=-frac{1}{2}int frac{d(1+t^2)}{1+t^2}=-frac{1}{2}cdot ln|1+t^2|+C=Big[; t=frac{1}{x}; Big]=\\=-frac{1}{2}cdot ln(1+frac{1}{x^2})+C=-frac{1}{2}cdot lnfrac{x^2+1}{x^2}+C=lnsqrt{frac{x^2}{x^2+1}}+C; ;$