Question

Найти неопределенный интеграл:

Answer 1

$\int x^2arctg(x)dx=(*)\\u=arctgx\\du=\frac{dx}{1+x^2} \\dv=x^2\\v=\frac{x^3}{3} \\(*)=\frac{x^3arctgx}{3} -\frac{1}{3} \int \frac{x^3}{1+x^2} dx=(*)\\\int \frac{x^3}{1+x^2}dx=\int \frac{x(1+x^2)-x}{1+x^2} dx=\int xdx-\int\frac{xdx}{1+x^2} =\frac{x^2}{2} -\frac{1}{2} \int \frac{d(1+x^2)}{1+x^2} =\\=\frac{x^2}{2}-\frac{1}{2} \ln(1+x^2)+c_1\\(*)=\frac{x^3arctgx}{3}+\frac{\ln(1+x^2)}{6} -\frac{x^2}{6} +c_2=\frac{1}{6} (2x^3arctgx+ \ln(1+x^2)-x^2+C)$