как взять интеграл от dx/(1+x^2)^2

нужно сделать замену
пусть t=arctg x, x=tg t
тогда dx/dt = 1/cos^2 t
dx = dt / cos^2 t
1+x^2=1+ tg^2 t = 1/cos^2 t
(1+x^2)^2 = 1/cos^4 t
1/(1+ x^2)^2 = cos^4 t
dx/(1+ x^2)^2 = cos^4 t / cos^2 t dt = cos^2 t dt = 1/2 (1+ cos(2t)) dt
1/2 dt = d(1/2 t)
1/2 cos(2t) dt = 1/4 cos(2t) d (2t) = 1/4 d(sin(2t)) = d (1/4 sin(2t))
dx/(1+ x^2)^2 = d(1/2 t) + d (1/4 sin(2t)) = d(1/2 t + 1/4 sin(2t))
1/4 sin 2t = 1/2 sin t cos t = 1/2 cos^2 t tg t = (1/2 tg t)/(1+ tg^2 t) = x / (2(1+x^2))
dx/(1+ x^2)^2 = d (1/2 arctg x + x / (2(1+x^2)))
Интеграл равен 1/2 arctg x + x / (2(1+x^2)) + С