"\\int_1^4 \\frac{1}{(x-3)^{\\frac{1}{3}}}dx =\n \\left[\n\u00a0\u00a0\\begin{array}{}\n\u00a0\u00a0\u00a0\u00a0\u00a0u = x -3 & 4 \\to1\\\\\n\u00a0\u00a0\u00a0\u00a0\u00a0du = dx & 1 \\to -2\n\u00a0\u00a0\\end{array}\n\\right]=\\\\\n=\\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 = \\\\\n=\\frac{3}{2}(1-(-2)^{\\frac{2}{3}}) = \\frac{3}{2} - \\frac{3}{\\sqrt[3]{2}}"
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