Question

Математика, найти первообразные

Answer 1

$4); ; int frac{dx}{arctgxcdot (x^2+1)} =int frac{frac{dx}{x^2+1}}{arctgx} =int frac{d(arctgx)}{arctgx}=Big [int frac{dt}{t}Big ]=ln|arctgx|+C\\5); ; int frac{dx}{x^2-6x+6} =int frac{dx}{(x-3)^2-3} =int frac{d(x-3)}{(x-3)^2-3} =Big [, int frac{dt}{t^2-a^2}Big ]= \\=frac{1}{2sqrt{3}}cdot lnBig |frac{x-3-sqrt{3}}{x-3+sqrt{3}} Big |+C$

$6); ; int frac{2x-3}{x^2-2x-3} , dx=int frac{2x-3}{(x-1)^2-4}, dx=Big [, t=x-1,; x=t+1,; dx=dtBig ]=\\= int frac{2t-1}{t^2-4}=int frac{2t, dt}{t^2-4}-int frac{dt}{t^2-4} =int frac{d(t^2-4)}{t^2-4}-frac{1}{2cdot 2} cdot lnBig | frac{t-2}{t+2}Big |=[int frac{du}{u}]=\\= ln|t^2-4|- frac{1}{4}cdot lnBig |frac{t-2}{t+2}Big |+C= ln|x^2-2x-3|- frac{1}{4}, lnBig |frac{x-3}{x+1} Big |+C$

$7); ; int (4-5x); e^{x}dx=[, u=4-5x,; du=-5, dx,; dv=e^{x}, dx,; v=e^{x}]=\\=uv-int v, du=(4-5x)e^{x}+5cdot int e^{x}, dx=\\=(4-5x), e^{x}+5e^{x}+C=e^{x}, (4-5x+5)+C=(9-5x), e^{x}+C$

Answer 2

Во 2 примере надо: (x-3)^2-3 , вместо (х-3)^2+3