Question

найти интеграл 2x arctgx dx

Answer 1

$\int udv=uv-\int v du\\\\\int 2x*arctgxdx=[u=arctgx\rightarrow du=\frac{dx}{1+x^2};dv=2xdx\rightarrow v=x^2]=\\=arctgx*x^2-\int\frac{x^2}{1+x^2}dx=[\int\frac{x^2}{x^2+1}dx=\int\frac{x^2+1-1}{x^2+1}dx=\\=\int(1-\frac{1}{x^2+1})dx=x-arctgx+C]=\\=arctgx*x^2-x+arctgx+C$