Question

数学积分题，求大神解答，感激不尽

Answer 1

原式=∫x(sec²x-1)dx
=∫xsec²xdx-∫xdx
=∫xdtanx-x²/2
=xtanx-∫tanxdx-x²/2
=xtanx-ln|secx|-x²/2+c

Answer 2

结果为:

((sin(2*x)^2+cos(2*x)^2+2*cos(2*x)+1)*log(sin(2*x)^2+cos(2*x)^2+2*cos(2*x)+1)-x^2*sin(2*x)^2+4*x*sin(2*x)-x^2*cos(2*x)^2-2*x^2*cos(2*x)-x^2)/(2*sin(2*x)^2+2*cos(2*x)^2+4*cos(2*x)+2)