Question

Помогите решить интеграл с 23-30

Answer 1

$displaystyleintfrac{2x-1}{3x^2-2x+6}dx=frac{1}{3}intfrac{6x-2-1}{3x^2-2x+6}dx=frac{1}{3}intfrac{d(3x^2-2x+6)}{3x^2-2x+6}-\-frac{1}{3sqrt3}intfrac{d(sqrt3x-frac{1}{sqrt3})}{(sqrt3x-frac{1}{sqrt3})^2+frac{17}{3}}=frac{1}{3}ln|3x^2-2x+6|-\-frac{1}{3sqrt{17}}arctgfrac{3x-1}{sqrt{17}}+C\(3x^2-2x+6)'=6x-2$

$displaystyleintfrac{x-1}{sqrt{3x^2-x+5}}dx=frac{1}{6}intfrac{6x-1-5}{sqrt{3x^2-x+5}}dx=frac{1}{6}intfrac{d(3x^2-x+5)}{sqrt{3x^2-x+5}}-\-frac{5}{6sqrt3}intfrac{d(sqrt3x-frac{1}{2sqrt3})}{sqrt{(sqrt3x-frac{1}{2sqrt3})^2+frac{59}{12}}}=frac{1}{3}sqrt{3x^2-x +5}-\-frac{5}{6sqrt3}ln|sqrt3x-frac{1}{2sqrt3}+sqrt{3x^2-x+5}|+C\(3x^2-x+5)'=6x-1$

$displaystyleintfrac{sqrt{x^2+4}}{x}dx=intfrac{t^2-4+4}{t^2-4}dt=int(1+frac{4}{t^2-4})dt=\=t+ln|frac{t-2}{t+2}|+C=sqrt{x^2+4}+ln|frac{sqrt{x^2+4}-2}{sqrt{x^2-4}+2}|+C\4+x^2=t^2\x^2=t^2-4\xdx=tdt= textgreater dx=frac{tdt}{x};frac{dx}{x}=frac{tdt}{t^2-4}$

$displaystyleintfrac{dx}{(x-1)sqrt{x^2-1}}=-frac{1}{2}intfrac{d(2t+1)}{sqrt{2t+1}}=-sqrt{2t+1}+C=\=-sqrt{frac{x+1}{x-1}}+C\x-1=frac{1}{t};x=frac{1+t}{t}\dx=-frac{dt}{t^2}$

$displaystyleintfrac{lnx}{x^2}dx=-frac{lnx}{x}+intfrac{dx}{x^2}=-frac{lnx}{x}-frac{1}{x}+C=-frac{1}{x}(lnx+1)+C\u=lnx;du=frac{dx}{x}\dv=frac{dx}{x^2};v=-frac{1}{x}$

$displaystyleint xarctg2xdx=frac{x^2}{2}arctg2x-frac{1}{4}intfrac{1+4x^2-1}{1+4x^2}dx=frac{x^2}{2}arctg2x-\-frac{1}{4}int(1-frac{1}{1+4x^2})dx=frac{x^2}{2}arctg2x-frac{x}{4}+frac{1}{8}arctg2x+C=\=frac{arctg2x}{2}(x^2+frac{1}{4})-frac{x}{4}+C\u=arctg2x;du=frac{2dx}{1+4x^2}\dv=xdx;v=frac{x^2}{2}$

$displaystyleint(x-7)cos2xdx=frac{x-7}{2}sin2x-frac{1}{2}int sin2xdx=\=frac{x-7}{2}sin2x+frac{1}{4}cos2x+C\u=x-7;du=dx\dv=cos2xdx;v=frac{1}{2}sin2x$

$displaystyleint x^2e^{-x}dx=-x^2e^{-x}+2int xe^{-x}dx=-x^2e^{-x}-2xe^{-x}+\+2int e^{-x}dx=-x^2e^{-x}-2xe^{-x}-2e^{-x}+C=-e^{-x}(x^2+2x+2)+C\u=x^2;du=2xdx\dv=e^{-x}dx;v=-e^{-x}\p=x;dp=dx\dv=e^{-x}dx;v=-e^{-x}$