arctan（x的平方）的积分

∫arctan(x^2)dx
=xarctan(x^2)-∫x*1/(1+x^4)*2xdx
=xarctan(x^2)-∫2x^2/(1+x^4)dx
考虑∫2x^2/(1+x^4)dx
=∫2/(x^2+1/x^2)dx
=∫1/(x^2+1/x^2)[d(x+1/x)+d(x-1/x)]
=∫1/(x^2+1/x^2)d(x+1/x)+∫1/(x^2+1/x^2)d(x-1/x)
=∫d(x+1/x)/[(x+1/x)^2-2]+∫d(x-1/x)/[(x-1/x)^2+2)
=∫[1/(x+1/x-√2)-1/(x+1/x+√2)]*1/(2√2)d(x+1/x)+√2/2*arctan[√2/2*(x-1/x)]
=√2/4*(ln|x+1/x-√2|-ln|x+1/x+√2|)+√2/2*arctan[√2/2*(x-1/x)]+C
=√2/4*ln|(x+1/x-√2)/(x+1/x+√2)|+√2/2*arctan[√2/2*(x-1/x)]+C
故∫arctan(x^2)dx
=xarctan(x^2)+√2/4*ln|(x+1/x+√2)/(x+1/x-√2)|-√2/2*arctan[√2/2*(x-1/x)]+C1

先分部积分，再用有理函数积分