Question

Найти интеграл ((x^4-2*x^3+3x+4)/(1+x^3)) dx

Answer 1

Ответ:

Пошаговое объяснение:

$\displaystyle \int \frac{x^4-2x^3+3x+4}{1+x^3} dx=\int\frac{2x+6x}{x^3+1} dx + \int xdx-2\int dx=$

$\displaystyle =2\int \frac{x+3}{(x+1)(x^2-x+1)} dx+\frac{x^2}{2} -2x=$

$\displaystyle = 2 \int \frac{2}{3(x+1)} dx-2 \int \frac{2x-7}{3(x^2-x+1)} +\frac{x^2}{2} -2x=$

$\displaystyle = \frac{4}{3} ln(x+1)-2\int \frac{2x-1}{x^2-x+1} dx-2\int\frac{6}{x^2+x+1} dx+\frac{x^2}{2} -2x=$

$\displaystyle =\frac{4}{3} ln(x+1) -2\left[\begin{array}{ccc}u=x^2-x+1\\dx=\displaystyle \frac{1}{2x-1} du\\\end{array}\right] -12\left[\begin{array}{ccc}v=\displaystyle \frac{2x-1}{\sqrt{3} } \\dx=\displaystyle \frac{\sqrt{3} }{2} dv\\\end{array}\right] +\frac{x^2}{2} -2x=$

$\displaystyle =\frac{4}{3} ln(x+1)-\frac{2}{3} ln(x^2-x+1)+\frac{8}{\sqrt{3} } arctg \bigg ( \frac{2x-1}{\sqrt{3} } \bigg ))+\frac{x^2}{2} -2x+C$