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Question 1

Question 2

Решите задачу:

$1)\; \; \int \frac{cos2x\, dx}{cosx-sinx}=\int \frac{(cos^2x-sin^2x)dx}{cosx-sinx}=\int \frac{(cosx-sinx)(cosx+sinx)}{cosx-sinx}dx=\\\\=\int (cosx+sinx)dx=sinx-cosx+C\\\\2)\; \; \int \frac{ln^4x}{x}dx=\int ln^4x\cdot \frac{dx}{x}}=[\frac{dx}{x}=d(lnx)]= \frac{ln^5x}{5}+C\\\\3)\; \; \int \frac{dx}{\sqrt{(1-x^2)arcsinx}}=\int \frac{1}{\sqrt{arcsinx}}\cdot \frac{dx}{\sqrt{1-x^2}} =[\frac{dx}{\sqrt{1-x^2}}=d(arcsinx)]=\\\\=\int \frac{d(arcsinx)}{\sqrt{arcsinx}} =2\sqrt{arcsinx}+C$

$4)\; \; \int \frac{x\, dx}{ \sqrt{x^2+4x+4} }=\int \frac{x\, dx}{\sqrt{(x+2)^2}}=\int \frac{x\, dx}{x+2}=\int (1-\frac{2}{x+2} )=\\\\=x-2\cdot ln|x+2|+C\\\\5)\; \; \int \frac{(2x-3)}{3x^2-7x+11}=\int \frac{(2x-3)dx}{3(x^2-\frac{7}{3}x+\frac{11}{3})}=\frac{1}{3}\int \frac{(2x-3)dx}{(x-\frac{7}{6})^2+\frac{83}{36}} =\\\\=[t=x-\frac{7}{6},\; dt=dx,\; x=t+\frac{7}{6} ]=\frac{1}{3}\int \frac{2t-\frac{2}{3}}{t^2+\frac{83}{36}}dt=\\\\=\frac{1}{3}\int \frac{2t\, dt}{t^2+\frac{83}{36}}-\frac{2}{9}\int \frac{dt}{t^2+\frac{83}{36}}-\frac{2}{9}\int \frac{dt}{t^2+\frac{83}{36}}=$

$= \frac{1}{3} \cdot ln|t^2+ \frac{83}{36}|-\frac{2}{9}\cdot \frac{6}{\sqrt{83}}\cdot arctg \frac{6t}{\sqrt{83}}+C=\\\\= \frac{1}{3} \cdot ln|(x-\frac{7}{6})^2+\frac{83}{36}|-\frac{4}{3\sqrt{83}} \cdot arctg \frac{6(x-\frac{7}{6})}{\sqrt{83}}+C$