画方框的不定积分怎样化解？

如下图，供参考。

积分函数 f(x) = sinxcosx/[(sinx)^4+(cosx)^4] = (1/2)sin2x/[1-2(sinxcosx)^2]
= (1/2)sin2x/[1-(1/2)(sin2x)^2] = sin2x/[2-(sin2x)^2]
= A/(√2-sin2x) + B/(√2+sin2x) = [√2(A+B)+(A-B)sin2x]/[2-(sin2x)^2]
A+B = 0, A-B = 1, 解得 A = 1/2, B = -1/2,
积分函数化为 f(x) = (1/2)[1/(√2-sin2x)-1/(√2+sin2x)]
I = (1/2) ∫ [1/(√2-sin2x)-1/(√2+sin2x)]dx
= (1/4) ∫ [1/(√2-sin2x)-1/(√2+sin2x)]d(2x)
令 tanx = u， 则 sin2x = 2/(1+u^2), d(2x) = 2du/(1+u^2)
I = (√2/4)∫du/[u^2-(√2-1)] - (√2/4)∫du/[u^2+(√2+1)]
= (√2/4)∫du/[u^2-(√2-1)] - (√2/4)∫du/[u^2+(√2+1)]
= {√2/[8√(√2-1)]} ∫{[1/[u-√(√2-1)]-[1/[u+√(√2-1)]}du - (√2/4)∫du/[u^2+(√2+1)]
= {√2/[8√(√2-1)]}ln|[u-√(√2-1)]/[u+√(√2-1)]| - {√2/[4√(√2+1)]}arctan[u/√(√2+1)] + C
= {√2/[8√(√2-1)]}ln|[tanx-√(√2-1)]/[tanx+√(√2-1)]|
- {√2/[4√(√2+1)]}arctan[tanx/√(√2+1)] + C

你能在糊涂一点吗？左边是sinx的4次吗？采用降幂