Question

Evaluate ∫_1^4▒1/〖(x-3)〗^(1/3) dx

Answer 1

$\int_1^4 \frac{1}{(x-3)^{\frac{1}{3}}}dx = \left[   \begin{array}{}      u = x -3 & 4 \to1\\      du = dx & 1 \to -2   \end{array} \right]=\\ =\int_{-2}^1 \frac{1}{u^{\frac{1}{3}}}du = \int_{-2}^1 u^{-\frac{1}{3}}du = \frac{u^{\frac{2}{3}}}{\frac{2}{3}}|_{-2}^1 = \\ =\frac{3}{2}(1-(-2)^{\frac{2}{3}}) = \frac{3}{2} - \frac{3}{\sqrt[3]{2}}$