Question

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Answer 1

Ответ:

$\displaystyle 1)\ \ \int \frac{(2x-1)(\sqrt{x}+\sqrt[7]{x})}{x^2}\, dx=\int \frac{2x\sqrt{x} +2x\sqrt[7]{x}-\sqrt{x}-\sqrt[7]{x}}{x^2}\, dx=\\\\\\=\int \frac{2x^{\frac{3}{2}}+2x^{\frac{8}{7}}-x^{\frac{1}{2}}-x^{\frac{1}{7}}}{x^2}\, dx=\int \Big(2x^{-\frac{1}{2}}+2x^{-\frac{6}{7}}-x^{-\frac{3}{2}}-x^{-\frac{13}{7}}\Big)\, dx=$

$\displaystyle =\frac{2\cdot 2x^{\frac{1}{2}}}{1}+\frac{2\cdot 7x^{\frac{1}{7}}}{1}-\frac{2x^{-\frac{1}{2}}}{-1}-\frac{7x^{-\frac{6}{7}}}{6}+C=4\sqrt{x}+14\sqrt[7]{x}+\frac{2}{\sqrt{x}}-\frac{7}{6\sqrt[7]{x^6}}$  

$2.\ a)\ \ \displaystyle \int\limits_0^{\pi }\, (sinx+1)\, dx=(-cosx+x)\Big|_0^{\pi }=-cos\pi +\pi -(-cos0+0)=\\\\=-(-1)+\pi +1=2+\pi \\\\\\b)\int\limits_1^9\, (2x^2+\sqrt{x}-1)\, dx=\Big(\frac{2x^3}{3}+\frac{2x^{\frac{3}{2}}}{3}-x\Big)\Big|_1^9=\\\\\\=486+\frac{2\sqrt{9^3}}{3}-9-\Big(\frac{2}{3}+\frac{2}{3}-1\Big)=486+18-9-\Big(\frac{4}{3}-1\Big)=495-\frac{1}{3}=494\frac{2}{3}$    

Answer 2

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