Задача 74631 Найти неопределенный интеграл...
Условие
Решение
u = x; dv = cos 3x dx; du = dx; v = 1/3*sin 3x
[m]\int u\ dv = uv - \int v\ du = \frac{x}{3} \sin\ 3x - \frac{1}{3}\int \sin\ 3x\ dx = [/m]
[m]= \frac{x}{3} \sin\ 3x - \frac{1}{3} \cdot \frac{1}{3} (-\cos\ 3x) + C = \frac{x}{3} \sin\ 3x + \frac{1}{9} \cos\ 3x + C [/m]
б) [m]\int (6 + 5x)\ \ln\ x\ dx[/m]
u = ln x; dv = (5x + 6) dx; du = 1/x dx; v = 5x^2/2 + 6x = 2,5x^2 + 6x
[m]\int u\ dv = uv - \int v\ du = (2,5x^2 + 6x)\ln\ x - \int \frac{2,5x^2 + 6x}{x}\ dx = [/m]
[m]= (2,5x^2 + 6x)\ln\ x - \int (2,5x + 6)\ dx = [/m]
[m] = (2,5x^2 + 6x)\ln\ x - 2,5 \cdot \frac{x^2}{2} - 6x + C= [/m]
[m] = (2,5x^2 + 6x)\ln\ x - 1,25x^2 - 6x + C[/m]