Математика: решить примеры, не понимаю...

решить примеры, не понимаю...

Ответ:

2313455345

12.10.2020 03:02

$\int {\frac{dx}{\sqrt{8-2x}} } = | 8-2x = t^2 = 2tdt = -2dx = dx = -tdt| = \int {\frac{-tdt}{\sqrt{t^2}} } = -\int {dt} } = t + c = \sqrt{8-2x} + c\\\int {\frac{6x-5}{3x^2-5x+4} } }dx = | 3x^2-5x+4 = t = dt = (6x-5)dx| = \int {\frac{dt}{t} } } = ln|t| + c = ln(3x^2-5x+4) + lnC = lnC(3x^2-5x+4)\\$

$\int {\frac{arcsin^3(2x)}{\sqrt{1-4x^2}} }dx = |arcsin(2x) = t = dt = \frac{2dx}{\sqrt{1-4x^2}} = \frac{dt}{2} = \frac{dx}{\sqrt{1-4x^2}}| = \int {\frac{t^3*\frac{dt}{2} }{1} } = \int{\frac{1}{2}t^3dt } = \frac{1}{8}t^4 + c =\frac{1}{8}arcsin^4(2x) + c\\$

$\int{x*arccos(2x)dx} = |arccos(2x) = u = du = -\frac{2dx}{\sqrt{1-4x^2}}. dv = xdx = v = \frac{1}{2}x^2| = \frac{1}{2}x^2*arccos(2x) + \int{\frac{2*\frac{1}{2}x^2}{\sqrt{1-4x^2}}dx = \frac{1}{2}x^2*arccos(2x) + \int{\frac{x^2}{\sqrt{1-4x^2}}dx$

$\int{\frac{x^2}{\sqrt{1-4x^2}}dx = | x = \frac{1}{2}sin(t) = dx = \frac{1}{2}cos(t)dt| = \int{\frac{0.25sin^2t * 0.5cos(t)}{cos(t)}dt = \frac{1}{8}\int{sin^2tdt} =$ $\frac{1}{16} \int{(1-cos2t})dt = \frac{1}{16}(t - \frac{1}{2}sin2t)+c = |x = 0.5sin(t) = t = arcsin(2x)| = \frac{1}{16}(arcsin(2x) - \frac{1}{2}sin(arcsin(2x))+c = \frac{1}{16}(arcsin(2x) - x) + c\\\int {x*arccos(2x)dx} = \frac{1}{2}x^2*arccos(2x) + \frac{1}{16}(arcsin(2x) - x) + c$

$\int{\frac{x}{x^3-x^2}dx = \int{\frac{1}{(x-1)(x+1)}}dx = \int{\frac{1}{2}(\frac{1}{x-1} - \frac{1}{x+1})}dx = \frac{1}{2}(ln|x-1| - ln|x+1|)+c = \frac{1}{2}ln|\frac{x-1}{x+1}| + c$ $\int{\frac{13}{(x^2+4)(x+3)}dx = | \frac{Ax+B}{x^2+4} + \frac{C}{x+3} = \frac{(Ax+B)(x+3) + Cx^2 + 4C}{(x+3)(x^2+4)} = \frac{Ax^2 + 3Ax+Bx+3B + Cx^2+4C}{(x+3)(x^2+4)} =$ $\frac{x^2(A+C)+x(3A+B) + 3B + 4C}{(x+3)(x^2+4)} = \left \{ {{A+C=0} \atop {3A+B=0\\\} \atop {3B+4C=13}} \right. = \left \{ {{A=-C} \atop {B = -3A = 3C\\\} \atop {9C+4C=13 = 13C = 13 = C = 1}} \right.$ $A = -1;\\B = 3;\\C = 1\\\frac{13}{(x^2+4)(x+3)} = \frac{3-x}{x^2+4} + \frac{1}{x+3}|\\ \int{\frac{13}{(x^2+4)(x+3)}dx} = \int{ \frac{3-x}{x^2+4} + \frac{1}{x+3}}dx = -\int{\frac{x-3}{x^2+4}dx + ln(x+3) = -\int{\frac{xdx}{x^2+4} + 3\int\frac{dx}{x^2+2^2} + ln(x+3)$ $-\frac{1}{2}ln(x^2+4) + \frac{3}{2}arctg\frac{x}{2} + ln(x+3) + c$