Question

∫arctanx/(x^2(1+x^2))dx

Answer 1

令x = tanz,dx = sec²z dz

∫ arctanx/[x²(1 + x²)] dx

= ∫ z/(tan²zsec²z) * (sec²z dz)

= ∫ zcot²z dz

= ∫ z(csc²z - 1) dz

= ∫ zcsc²z dz - ∫ z dz

= ∫ z d(- cotz) - ∫ z dz

= - zcotz + ∫ cotz dz - ∫ z dz

= - zcotz + ln|sinx| - z²/2 + C

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