Question

Найдите неопределенные интегралы.

Answer 1

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