Question

Найти подстановкой
$\int\limits^1_0 { \frac{dx}{ (3x+1)^{4} } }$

Answer 1

$3x+1=t\\ 3dx=dt \\ \int\limits^1_0 {\dfrac{dt}{3t^4}} = |^{1}_{0} \ -\frac{1}{9} \cdot t^{-3} = \frac{-1}{9t^3} = |^{1}_{0} = - \frac{1}{9(3x+1)^3} = -\frac{1}{576} + \frac{1}{9} = \frac{7}{64}$