arctanx/(x^2)的原函数
2个回答

∫ arctanx / x² dx

= -∫ arctanx d(1/x)

= -(1/x)arctanx + ∫ (1/x)d(arctanx)

= -(1/x)arctanx + ∫ 1/[x(1+x²)] dx

= -(1/x)arctanx + ∫ (1/x)dx - ∫ x/(1+x²) dx

= -(1/x)arctanx + ln|x| - (1/2)ln(1+x²) + C

Note:

1/[x(1+x²)] = A/x + (Bx+C)/(1+x²)

1 = A(1+x²) + (Bx+C)x

1 = Ax²+A+Bx²+Cx

1 = (A+B)x²+Cx+A

1 = A

A+B = 0 => B=-1

C = 0

1/[x(1+x²)] = 1/x - x/(1+x²)