Question

СРОЧНО ПОМОГИТЕ ПРОШУ ВАС, СПАСИБО

Answer 1

$1)\; \; \int \frac{1-\sqrt[6]{x}+2\sqrt[3]{x}}{x+2\sqrt{x^3}+\sqrt[3]{x^4}} =\Big [x=t^6\; ,\; dx=6t^5\, dt\Big ]=\int \frac{1-t+2t^2}{t^6+2t^9+t^{8}} \cdot 6t^5dt=\\\\=6\cdot \int \frac{2t^2-t+1}{t^5(2t^4+t^3+t)} \cdot t^5\, dt =6\cdot \int \frac{2t^2-t+1}{t\cdot (2t^3+t^2+1)}dt=6\cdot \int \frac{2t^2-t+1}{t\cdot (t+1)(2t^2-t+1)} dt=\\\\=6\cdot \int \frac{dt}{t(t+1)} =6\cdot \int ( \frac{1}{t}- \frac{1}{t+1} )dt=6\cdot (ln|t|-ln|t+1|)+C=\\\\=6\cdot (ln|\sqrt[6]{x}|-ln|\sqrt[6]{x}+1|)+C$

$2)\; \; \int \frac{dx}{cos^4x} =\int \frac{1}{cos^2x} \cdot \frac{dx}{cos^2x} =\int (1+tg^2x)\cdot \frac{dx}{cos^2x}=\\\\=[t=tgx,\; dt=\frac{dx}{cos^2x}]=\int (1+t^2)dt=t+\frac{t^3}{3}+C=tgx+\frac{tg^3x}{3}+C$

$3)\; \; \int \frac{cosx\cdot dx}{2+cosx} =\Big [t=tg\frac{x}{2}\; ,\; cosx=\frac{1-t^2}{1+t^2}\; ,\; dx=\frac{2dt}{1+t^2}\Big ]=\\\\=2\cdot \int \frac{(1-t^2)dt}{(1+t^2)(2+\frac{1-t^2}{1+t^2})} =-2\cdot \int \frac{(t^2-1)dt}{t^2+3} =-2\cdot \int (1-\frac{4}{t^2+3})dt=\\\\=-2\cdot (t-4\cdot \frac{1}{\sqrt3}\cdot arctg\frac{t}{\sqrt3})+C=-2tg\frac{x}{2}+\frac{8}{\sqrt3}\cdot arctg(\frac{tg\frac{x}{2}}{\sqrt3})+C$

$4)\; \; \int \frac{2ctgx+1}{(2sinx+cosx)^2} dx=\int \frac{\frac{2}{tgx}+1}{4sin^2x+4sinx\cdot cosx+cos^2x} dx=\Big [\frac{:cos^2x}{:cos^2x}\Big ]=\\\\=\int \frac{(2+tgx)\cdot \frac{dx}{cos^2x}}{tgx\cdot (4tg^2x+4tgx+1)} =[\; t=tgx\; ]=\int \frac{(2+t)dt}{t\cdot (4t^2+4t+1)} =\int \frac{(2+t)dt}{t(2t+1)^2}=I \\\\ \frac{2+t}{t(2t+1)^2} =\frac{A}{t}+\frac{B}{(2t+1)^2}+\frac{C}{2t+1} = \frac{A(2t+1)^2+Bt+Ct(2t+1)}{t(2t+1)^2} ;\\\\2+t=A(2t+1)^2+Bt+Ct(2t+1)\\\\2+t=4At^2+4At+A+Bt+2Ct^2+Ct$

$t^2\; \; |\; 4A+2C=0\\\\t\; \; |\; 4A+B+C=1\\\\t^0\; \; |\; A=2\\\\C=-2A=-4\; ,\; \; B=1-C-4A=1+4-8=-3\\\\I=\int \frac{2+t}{t(2t+1)^2} dt=\int ( \frac{2}{t}-\frac{3}{(2t+1)^2}-\frac{4}{2t+1} )dt=\\\\=2ln|t|+3\cdot \frac{1}{2t+1}-2ln|2t+1|+C=\\\\=2ln|tgx|+\frac{3}{2tgx+1}-2ln|2tgx+1|+C$