Question

不定积分 x (x^1\2-1) ^4 dx

Answer 1

如果题目是求不定积分：∫[x((√x)-1)^4]dx=∫[x(x+1-2√x)²]dx=∫[x(x²+1+6x-4x^(3/2)-4√x)]dx=∫[(x³+x+6x²-4x^(5/2)-4x^(3/2))]dx=0.25x^4+0.5x²+2x³-(8/7)*x^(7/2)-(8/5)*x^(5/2)+c（c为任意常数）

Answer 2

答案：∫/(x^-)dx=/[∫/(x-)dx-∫/(x+)dx]=/[ln(x--ln(x+)+c]=/ln

追问

原来如此，那接下来呢？