Задача 75887 решите первые 3 интеграла...
Условие
математика ВУЗ
53
Решение
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Замена y = 2x^2 + 3, dy = 4x dx; x dx = 1/4*dy
[m]\int \frac{1}{4} \cdot \frac{dy}{\sqrt{y}} = \frac{1}{4} \int y^{-1/2}dy =[/m]
[m]= \frac{1}{4} \frac{y^{1/2}}{1/2} +C = \frac{2}{4} y^{1/2} + C = \frac{1}{2} \sqrt{2x^2+3}+C[/m]
2) [m]\int e^{-4x^3} x^2 dx[/m]
Замена y = -4x^3; dy = -12x^2 dx; x^2 dx = -1/12 dy
[m]\int (-\frac{1}{12} \cdot e^y dy) = -\frac{1}{12} \cdot e^y + C = -\frac{1}{12} \cdot e^{-4x^3} + C[/m]
3) [m]\int \frac{1 + \ln(x-1)}{x-1} dx[/m]
Замена y = 1 + ln(x - 1); dy = dx/(x - 1)
[m]\int y\ dy = \frac{y^2}{2} + C= \frac{(1 + ln(x - 1))^2}{2} + C[/m]