Question

高数：∫（0→1）xarctanx/(1+x^2)^3 dx

Answer 1

令x = tanz，dx = sec²z dz
∫(0→1) xarctanx/(1 + x²)³ dx
= ∫(0→π/4) ztanz/sec⁶z * (sec²z dz)
= ∫(0→π/4) zsinzcos³z dz
= ∫(0→π/4) zcos³z d(- cosz)
= (- 1/4)∫(0→π/4) z d(cos⁴z)
= (- 1/4)zcos⁴z |(0→π/4) + (1/4)∫(0→π/4) cos⁴z dz
= - π/64 + (1/4)∫(0→π/4) [(1 + cos2z)/2]² dz
= - π/64 + (1/4)²∫(0→π/4) [1 + 2cos2z + (1 + cos4z)/2] dz
= - π/64 + (1/16)[z + sin2z + z/2 + (1/8)sin4z] |(0→π/4)
= - π/64 + (1/128)(8 + 3π)
= (8 + π)/128

Answer 2

令 arctanx=t 带入得∫tant *t/(1+tant^2)^2 dt
1+tant^2=1/(cost^2) 原式=∫tant *t*cost^2=(∫sin2t t dt)/2 剩下的自己分部积分吧。。

Answer 3

令x=tant，则arctanx=t
然后按照常规的求算就可以了