Математика: Решить неоприделений интеграл dx/((x+1)^2*(x^3+4))

Решить неоприделений интеграл dx/((x+1)^2*(x^3+4))

Ответ:

oleg34643

03.08.2020 14:54

В общем либо интеграл с ошибкой, либо опечатка. Касается это дело х³+4. $\int\frac{dx}{(x+1)^2(x^3+4)}\\\frac{1}{(x+1)^2(x^3+4)}=\frac{A}{x+1}+\frac{B}{(x+1^2)}+\frac{Cx^2+Dx+E}{x^3+4}\\1=A(x^3+4)(x+1)+B(x^3+4)+(Cx^2+Dx+E)(x+1)^2\\1=A(x^4+x^3+4x+1)+B(x^3+4)+C(x^4+2x^3+x^2)+\\+D(x^3+2x^2+x)+E(x^2+2x+1)\\x^0|1=4A+4B+E\\x^1|0=4A+D+2E\\x^2|0=C+2D+E\\x^3|0=A+B+2C+D\\x^4|0=A+C$ $\left[\begin{array}{ccccc|c|c}4&4&0&0&1&1&-4\\4&0&0&1&2&0&-4\\0&0&1&2&1&0&\\1&1&2&1&0&0&-1\\1&0&1&0&0&0&\uparrow\end{array}\right]=\ \textgreater \ \left[\begin{array}{ccccc|c|c}0&4&-4&0&1&1&-4\\0&0&-4&1&2&0&\\0&0&1&2&1&0&\\0&1&1&1&0&0&\uparrow\\1&0&1&0&0&0&\end{array}\right]=\ \textgreater \ \\=\ \textgreater \ \left[\begin{array}{ccccc|c|c}0&0&-8&-4&1&1&8\\0&0&-4&1&2&0&4\\0&0&1&2&1&0&\uparrow\\0&1&1&1&0&0&\\1&0&1&0&0&0&\end{array}\right]=\ \textgreater \$ $=\ \textgreater \ \left[\begin{array}{ccccc|c|c}0&0&0&12&9&1&\\0&0&0&9&6&0&:9\\0&0&1&2&1&0&\\0&1&1&1&0&0&\\1&0&1&0&0&0&\end{array}\right]=\ \textgreater \ \left[\begin{array}{ccccc|c|c}0&0&0&12&9&1&-12\\0&0&0&1&\frac{2}{3}&0&\uparrow\\0&0&1&2&1&0&\\0&1&1&1&0&0&\\1&0&1&0&0&0&\end{array}\right]\\=\ \textgreater \ \left[\begin{array}{ccccc|c|c}0&0&0&0&1&1&\\0&0&0&1&\frac{2}{3}&0&\\0&0&1&2&1&0&\downarrow\\0&1&1&1&0&0&-1\\1&0&1&0&0&0&-1\end{array}\right]=\ \textgreater \$ $\left[\begin{array}{ccccc|c|c}0&0&0&0&1&1&\\0&0&0&1&\frac{2}{3}&0&\downarrow\\0&0&1&2&1&0&-2\\0&1&0&-1&-1&0&1\\1&0&0&-2&-1&0&2\end{array}\right]=\ \textgreater \ \left[\begin{array}{ccccc|c|c}0&0&0&0&1&1&\downarrow\\0&0&0&1&\frac{2}{3}&0&-\frac{2}{3}\\0&0&1&0&-\frac{1}{3}&0&\frac{1}{3}\\0&1&0&0&-\frac{1}{3}&0&\frac{1}{3}\\1&0&0&0&\frac{1}{3}&0&-\frac{1}{3}\end{array}\right]\\$ $\left[\begin{array}{ccccc|c|c}0&0&0&0&1&1&\\0&0&0&1&0&-\frac{2}{3}&\\0&0&1&0&0&\frac{1}{3}&\\0&1&0&0&0&\frac{1}{3}&\\1&0&0&0&0&-\frac{1}{3}&\end{array}\right]\\A=-\frac{1}{3};B=\frac{1}{3};C=\frac{1}{3};D=-\frac{2}{3};E=1$ $\frac{1}{(x+1)^2(x^3+4)}=-\frac{1}{3(x+1)}+\frac{1}{3(x+1)^2}+\frac{\frac{1}{3}(x^2-2x+3)}{x^3+4}$
$\frac{(x^2-2x+3)}{x^3+4}=\frac{A}{x+4^\frac{1}{3}}+\frac{Bx+C}{x^2-4^\frac{1}{3}x+4^\frac{2}{3}}\\(x^2-2x+3)=A(x^2-4^\frac{1}{3}x+4^\frac{2}{3})+B(x^2+4^\frac{1}{3}x)+C(x+4^\frac{1}{3})\\x^0|3=4^\frac{2}{3}A+4^\frac{1}{3}C\\x^1|-2=-4^\frac{1}{3}A+4^\frac{1}{3}B+C\\x^2|1=A+B=\ \textgreater \ B=1-A\\\begin{cases}3=4^\frac{2}{3}A+4^\frac{1}{3}C\\-\\-2=-4^\frac{1}{3}A+4^\frac{1}{3}(1-A)+C|*4^\frac{1}{3}\end{cases}\\3+2*4^\frac{1}{3}=3*4^\frac{2}{3}A-4^\frac{2}{3}\\A=\frac{3+2*4^\frac{1}{3}+4^\frac{2}{3}}{3*4^\frac{2}{3}}$
$B=1-\frac{3+2*4^\frac{1}{3}+4^\frac{2}{3}}{3*4^\frac{2}{3}}=\frac{-3-2*4^\frac{1}{3}+2*4^\frac{2}{3}}{3*4^\frac{2}{3}}\\3=\frac{3+2*4^\frac{1}{3}+4^\frac{2}{3}}{3}+4^\frac{1}{3}C\\\\C=\frac{6-2*4^\frac{1}{3}-4^\frac{2}{3}}{3*4^\frac{1}{3}}\\\frac{(x^2-2x+3)}{x^3+4}=\frac{\frac{3+2*4^\frac{1}{3}+4^\frac{2}{3}}{3*4^\frac{2}{3}}}{x+4^\frac{1}{3}}+\frac{\frac{-3-2*4^\frac{1}{3}+2*4^\frac{2}{3}}{3*4^\frac{2}{3}}x+\frac{6-2*4^\frac{1}{3}-4^\frac{2}{3}}{3*4^\frac{1}{3}}}{x^2-4^\frac{1}{3}x+4^\frac{2}{3}}$
$\int\frac{dx}{(x+1)^2(x^3+4)}=-\frac{1}{3}\int\frac{d(x+1)}{(x+1)}+\frac{1}{3}\int\frac{d(x+1)}{(x+1)^2}+\frac{1}{3}\int\frac{(x^2-2x+3)}{(x^3+4)}=\\=-\frac{1}{3}ln|x+1|+(x+1)^3+\frac{1}{3}\int\frac{(\frac{3+2*4^\frac{1}{3}+4^\frac{2}{3}}{3*4^\frac{2}{3}})d(x+4^\frac{1}{3})}{x+4^\frac{1}{3}}+\\+\frac{1}{3}\int\frac{(\frac{-3-2*4^\frac{1}{3}+2*4^\frac{2}{3}}{3*4^\frac{2}{3}}x+\frac{6-2*4^\frac{1}{3}-4^\frac{2}{3}}{3*4^\frac{1}{3}})dx}{x^2-4^\frac{1}{3}x+4^\frac{2}{3}}=$
$=-\frac{1}{3}ln|x+1|+(x+1)^3+(\frac{3+2*4^\frac{1}{3}+4^\frac{2}{3}}{9*4^\frac{2}{3}})ln|x+4^\frac{1}{3}|-\\-\frac{3+2*4^\frac{1}{3}-2*4^\frac{2}{3}}{18*4^\frac{2}{3}}*ln|x^2-4^\frac{1}{3}x+4^\frac{2}{3}|+\frac{3-2*4^\frac{1}{3}}{3\sqrt3*4^\frac{2}{3}}arctg\frac{2x-4^\frac{1}{3}}{\sqrt3*4^\frac{1}{3}}+C$