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Решите задачу:

$\int\limits^1_0 {(\frac{6}{\sqrt{3x+1}}-x)} \, dx = 6*\int\limits^1_0 {\frac{1}{\sqrt{3x+1}}} \, dx - \int\limits^1_0 {x} \, dx =\\\\ = 6*\frac{1}{3}*\int\limits^1_0 {\frac{1}{\sqrt{3x+1}}} \, d(3x) - \frac{x^2|^1_0}{2}=\\\\ = 6*\frac{1}{3}*\int\limits^1_0 {\frac{1}{\sqrt{3x+1}}} \, d(3x+1) - \frac{1^2-0^2}{2}=\\\\ = 6*\frac{1}{3}*\int\limits^{3*1+1}_{3*0+1} {\frac{1}{\sqrt{t}}} \, dt - 0.5=\\\\ = 2*\int\limits^{4}_{1} {t^{-\frac{1}{2}}}} \, dt - 0.5=\\\\$

$=2*\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}|^4_1-0.5=\\\\= 2*\frac{t^{\frac{1}{2}}}{\frac{1}{2}}|^4_1- 0.5=\\\\ =4*\sqrt{t}|^4_1- 0.5=\\\\ =4*(\sqrt{4}-\sqrt{1})-0.5=\\\\ =4*1-0.5=3.5$