∫（x^3-1）∕(4x^3-x)d(x)
=∫（x^3-x/4+x/4-1）∕(4x^3-x)d(x)
= 1/4+ ∫ (x/4-1)∕(4x^3-x) d(x)
=1/4+ ∫ {-9/(8(2x+1)) - 7/(8(2x-1)) + 1/x}dx
=1/4-(9ln|2x+1|)/16-(7ln|2x-1|)/16+ln|x|+C
不定积分∫（x^3-1）∕(4x^3-x)d(x)怎么做各位帮帮忙吧,谢谢了!要求详细的过程.
∫（x^3-1）∕(4x^3-x)d(x)
=∫（x^3-x/4+x/4-1）∕(4x^3-x)d(x)
= 1/4+ ∫ (x/4-1)∕(4x^3-x) d(x)
=1/4+ ∫ {-9/(8(2x+1)) - 7/(8(2x-1)) + 1/x}dx
=1/4-(9ln|2x+1|)/16-(7ln|2x-1|)/16+ln|x|+C