题目

高数：∫（0→1）xarctanx/(1+x^2)^3 dx

提问时间：2021-03-22

答案

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