问题描述: 高数:∫(0→1)xarctanx/(1+x^2)^3 dx 1个回答 分类:数学 2014-10-24 问题解答: 我来补答 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 展开全文阅读