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Question 1

Question 2

$\int\frac{3xdx}{2x^2-5}=\frac{3}{4}\int\frac{d(2x^2-5)}{2x^2-5}=\frac{3}{4}ln|2x^2-5|+C$
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$\int\frac{dx}{\sqrt{x^2-x+3}}=\int\frac{dx}{\sqrt{x^2-x+\frac{1}{4}+\frac{11}{4}}}=\int\frac{d(x-\frac{1}{2})}{\sqrt{(x-\frac{1}{2})^2+\frac{11}{4}}}=\\=ln|x-\frac{1}{2}+\sqrt{(x-\frac{1}{2})^2+\frac{11}{4}}}|+C$
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$\int(3x+1)cos2xdx=\frac{1}{2}(3x+1)sin2x-\frac{3}{2}\int sin2x=\\=\frac{1}{2}(3x+1)sin2x+\frac{3}{4}cos2x+C\\\\u=3x+1=\ \textgreater \ du=3dx\\dv=cos2x=\ \textgreater \ v=\frac{1}{2}sin2x$
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$\int\frac{xdx}{\sqrt{9-x^4}}=\frac{1}{2}\int\frac{d(x^2)}{\sqrt{9-x^4}}=\frac{1}{2}arcsin\frac{x^2}{3}+C$
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$\int\frac{\sqrt{x}dx}{1+\sqrt[4]{x}}=\int\frac{4t^5dt}{1+t}=4\int (t^4-t^3+t^2-t-\frac{1}{t+1}+1)dt=\\=4(\frac{t^5}{5}-\frac{t^4}{4}+\frac{t^3}{3}-\frac{t^2}{2}-ln|t+1|+t)+C=\\=\frac{4x\sqrt[4]{x}}{5}-x+\frac{4\sqrt[4]{x^3}}{3}-2\sqrt{x}-ln|\sqrt[4]{x}+1|+\sqrt[4]{x}+C\\\\\sqrt[4]{x}=t=\ \textgreater \ x=t^4;\sqrt{x}=t^2;dx=4t^3dt$
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$\int\frac{sin^3xdx}{1-cos2x}=\int\frac{sin^3xdx}{1-cos^2x+sin^2x}=\int\frac{sin^3xdx}{2sin^2x}=\frac{1}{2}\int sinxdx=\\=-\frac{1}{2}cosx+C$
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$\int\frac{(x+2)dx}{x^2+x-12}=\int\frac{(x+2)dx}{(x-3)(x+4)}=\int(\frac{5}{7(x-3)}+\frac{2}{7(x+4)})dx=\\=\frac{5}{7}ln|x-3|+\frac{2}{7}ln|x+4|+C\\\\\frac{x+2}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}=\frac{5}{7(x-3)}+\frac{2}{7(x+4)}\\x+2=A(x+4)+B(x-3)\\x|1=A+B=\ \textgreater \ A=1-B\\x^0|2=4A-3B\\2=4-4B-3B\\-2=-7B\\B=\frac{2}{7}\\A=1-\frac{2}{7}=\frac{5}{7}$
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$\int\frac{(x-3)dx}{x^2(x-1)}=\int(\frac{2}{x}+\frac{3}{x^2}-\frac{2}{x-1})dx=2ln|x|-\frac{3}{x}-2ln|x-1|+C=\\=2ln|\frac{x}{x-1}|-\frac{3}{x}+C\\\\\frac{x-3}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}=\frac{2}{x}+\frac{3}{x^2}-\frac{2}{x-1}\\x-3=A(x^2-x)+B(x-1)+Cx^2\\x^2|0=A+C=\ \textgreater \ C=-2\\x|1=-A+B=\ \textgreater \ A=2\\x^0|-3=-B=\ \textgreater \ B=3$

Alexаndr_zn · Answer 1 · 2018-05-26T16:26:55+0000

$\int\frac{3xdx}{2x^2-5}=\frac{3}{4}\int\frac{d(2x^2-5)}{2x^2-5}=\frac{3}{4}ln|2x^2-5|+C$
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$\int\frac{dx}{\sqrt{x^2-x+3}}=\int\frac{dx}{\sqrt{x^2-x+\frac{1}{4}+\frac{11}{4}}}=\int\frac{d(x-\frac{1}{2})}{\sqrt{(x-\frac{1}{2})^2+\frac{11}{4}}}=\\=ln|x-\frac{1}{2}+\sqrt{(x-\frac{1}{2})^2+\frac{11}{4}}}|+C$
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$\int(3x+1)cos2xdx=\frac{1}{2}(3x+1)sin2x-\frac{3}{2}\int sin2x=\\=\frac{1}{2}(3x+1)sin2x+\frac{3}{4}cos2x+C\\\\u=3x+1=\ \textgreater \ du=3dx\\dv=cos2x=\ \textgreater \ v=\frac{1}{2}sin2x$
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$\int\frac{xdx}{\sqrt{9-x^4}}=\frac{1}{2}\int\frac{d(x^2)}{\sqrt{9-x^4}}=\frac{1}{2}arcsin\frac{x^2}{3}+C$
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$\int\frac{\sqrt{x}dx}{1+\sqrt[4]{x}}=\int\frac{4t^5dt}{1+t}=4\int (t^4-t^3+t^2-t-\frac{1}{t+1}+1)dt=\\=4(\frac{t^5}{5}-\frac{t^4}{4}+\frac{t^3}{3}-\frac{t^2}{2}-ln|t+1|+t)+C=\\=\frac{4x\sqrt[4]{x}}{5}-x+\frac{4\sqrt[4]{x^3}}{3}-2\sqrt{x}-ln|\sqrt[4]{x}+1|+\sqrt[4]{x}+C\\\\\sqrt[4]{x}=t=\ \textgreater \ x=t^4;\sqrt{x}=t^2;dx=4t^3dt$
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$\int\frac{sin^3xdx}{1-cos2x}=\int\frac{sin^3xdx}{1-cos^2x+sin^2x}=\int\frac{sin^3xdx}{2sin^2x}=\frac{1}{2}\int sinxdx=\\=-\frac{1}{2}cosx+C$
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$\int\frac{(x+2)dx}{x^2+x-12}=\int\frac{(x+2)dx}{(x-3)(x+4)}=\int(\frac{5}{7(x-3)}+\frac{2}{7(x+4)})dx=\\=\frac{5}{7}ln|x-3|+\frac{2}{7}ln|x+4|+C\\\\\frac{x+2}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}=\frac{5}{7(x-3)}+\frac{2}{7(x+4)}\\x+2=A(x+4)+B(x-3)\\x|1=A+B=\ \textgreater \ A=1-B\\x^0|2=4A-3B\\2=4-4B-3B\\-2=-7B\\B=\frac{2}{7}\\A=1-\frac{2}{7}=\frac{5}{7}$
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$\int\frac{(x-3)dx}{x^2(x-1)}=\int(\frac{2}{x}+\frac{3}{x^2}-\frac{2}{x-1})dx=2ln|x|-\frac{3}{x}-2ln|x-1|+C=\\=2ln|\frac{x}{x-1}|-\frac{3}{x}+C\\\\\frac{x-3}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}=\frac{2}{x}+\frac{3}{x^2}-\frac{2}{x-1}\\x-3=A(x^2-x)+B(x-1)+Cx^2\\x^2|0=A+C=\ \textgreater \ C=-2\\x|1=-A+B=\ \textgreater \ A=2\\x^0|-3=-B=\ \textgreater \ B=3$