Question

求不定积分（（x-1)/(x+1))^4

Answer 1

这道绝非楼上想的这么简单。
∫ [(x - 1)/(x + 1)]⁴ dx
= ∫ [(x + 1) - 2]⁴/(x + 1)⁴ dx，运用二项式定理展开分子
= ∫ [(x + 1)⁴ - 4(x + 1)³(2) + 6(x + 1)²(2)² - 4(x + 1)(2)³ + 2⁴]/(x + 1)⁴ dx
= ∫ [1 - 8/(x + 1) + 24/(x + 1)² - 32/(x + 1)³ + 16/(x + 1)⁴] dx
= x - 8ln|x + 1| - 24/(x + 1) + 16/(x + 1)² - 16/[3(x + 1)³] + C
= - [8(9x² + 12x + 5)]/[3(x + 1)³] + x - 8ln|x + 1| + C

Answer 2

∫（x-1)/(x+1)^4 dx=∫（x+1-2)/(x+1)^4dx=∫[1/(x+1)^3-2/(x+1)^4]dx
=(- 1/2) 1/(x+1)² + (2/3) 1/(x+1)³ + C

追问

是（x-1)/(x+1)作为一个整体，然后再四次方~~麻烦再帮忙解答一下~~

追答

∫[（x-1)/(x+1)]^4 dx=∫[（x+1-2)/(x+1)]^4dx=∫[1 - 2/(x+1)]^4]dx
=∫[1 - 8/(x+1) + 24/(x+1)² - 32/(x+1)³ + 16/(x+1)^4]dx
=x - 8ln|x+1| - 24/(x+1) + 16/(x+1)² - 16/[3(x+1)³] + C