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Question 1

Question 2

Решите задачу:

$1)\; \; \int \frac{sinx\, dx}{\sqrt[3]{3+2cosx}}=[\, t=3+2cosx\; ,\; dt=-2sinx\, dx\, ]=\\\\=-\frac{1}{2}\int \frac{dt}{\sqrt[3]{t}} =- \frac{1}{2}\cdot \frac{t^{\frac{2}{3}}}{2/3}+C=- \frac{3}{4}\cdot \sqrt[3]{(3+2cosx)^2} +C\; ;\\\\(-\frac{3}{4}\cdot \sqrt[3]{(3+2cosx)^2}+C)'=-\frac{3}{4}\cdot \frac{2}{3}\cdot (3+2cosx)^{-\frac{1}{3}}\cdot (-2sinx)+0=\\\\=- \frac{1}{2}\cdot \frac{-2sinx}{\sqrt[3]{3+2cosx}}= \frac{sinx}{ \sqrt[3]{3+2cosx} } \; .$

$2)\; \; \int \frac{2x^2-3x+1}{x^3+1} =J\\\\ \frac{2x^2-3x+1}{(x+1)(x^2-x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\\\\2x^2-3x+1=A(x^2-x+1)+(Bx+C)(x+1)\\\\x=-1:\; \; A= \frac{2+3+1}{1+1+1}=2\\\\x^2\, |\; 2=A+B\; ,\; \; B=2-A=2-2=0\\\\x\; |\; -3=-A+B+C\\\\x^0\, |\; 1=A+C\; ,\; \; C=1-A=1-2=-1\\\\J=\int (\frac{2}{x+1}-\frac{1}{x^2-x+1} )dx=2\int \frac{dx}{x+1}-\int \frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}}=\\\\=2ln|x+1|-\frac{2}{\sqrt3}\cdot arctg \frac{2(x-\frac{1}{2})}{\sqrt3}+C=\\\\=2ln|x+1|-\frac{2}{\sqrt3}\cdot arctg\frac{2x-1}{\sqrt3}+C$

$3)\; \; \int \frac{(\sqrt[4]{x}+1)dx}{(\sqrt{x}+4)\sqrt[4]{x^3}}=[\, x=t^4\; ,\; dx=4t^3\, dt,\; t=\sqrt[4]{x}\, ]=\\\\=\int \frac{(t+1)\cdot 4t^3\, dt}{(t^2+4)\cdot t^3}=\int \frac{4(t+1)\, dt}{t^2+4}=2\cdot \int \frac{2t\, dt}{t^2+4}+4\cdot \int \frac{dt}{t^2+4}=\\\\=\{\int \frac{du}{u}=\ln|u|+C\}=2\cdot ln|t^2+4|+4\cdot \frac{1}{2}\cdot arctg\frac{t}{2}+C=\\\\=2\cdot ln(\sqrt{x}+4)+2arctg \frac{\sqrt{x}}{2}+C$