Вопрос в картинках...

Решите задачу:

$\int\limits^0_1 \, dx /{(6x-1)^4} = \int\limits^0_1 \, (6x-1)^{-4}dx = \frac{1}{6} \int\limits^0_1 \, (6x-1)^{-4}d(6x-1)= \\ = \frac{1}{6} \frac{(6x-1)^{-3}}{-3}|^0_1 = - \frac{1}{18} \frac{1}{(6x-1)^3}|^0_1 = - \frac{1}{18}( \frac{1}{(6*0-1)^3}- \frac{1}{(6*1-1)^3})= \\ =- \frac{1}{18}( \frac{1}{(-1)^3}- \frac{1}{5^3})= - \frac{1}{18}( -1 - \frac{1}{125})= - \frac{1}{18}*( - \frac{126}{125})= \frac{126}{2250} = \frac{7}{125}$