Question

Помогите решить интеграл с 23-30

Answer 1

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

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

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

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

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

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

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

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