∫x的八次/{x的平方+1}dxrt

x^8-1=(x^4-1)(x^4+1)
=(x²+1)(x²-1)(x^4+1)
所以
原式=∫(x^8-1+1)/(x²+1)dx
=∫(x²-1)(x^4+1)dx+∫1/(x²+1)dx
=∫(x^6-x^4+x²-1)dx+arctanx
=1/7x^7-1/5x^5+1/3x³-x+arctanx+c

x⁸ = x⁶[(x² + 1) - 1] = x⁶(x² + 1) - x⁶
= x⁶(x² + 1) - x⁴[(x² + 1) - 1] = x⁶(x² + 1) - x⁴(x² + 1) + x⁴
= x⁶(x² + 1) - x⁴(x² + 1) + x²[(x² + 1) - 1] = x⁶(x² + 1) - x⁴(x² + 1) + x²(x² + 1) - x²
= x⁶(x² + 1) - x⁴(x² + 1) + x²(x² + 1) - [(x² + 1) - 1]
= x⁶(x² + 1) - x⁴(x² + 1) + x²(x² + 1) - (x² + 1) + 1
∫ x⁸/(x² + 1) dx
= ∫ [x⁶(x² + 1) - x⁴(x² + 1) + x²(x² + 1) - (x² + 1) + 1]/(x² + 1) dx
= ∫ [x⁶ - x⁴ + x² - 1 + 1/(x² + 1)] dx
= x⁷/7 - x⁵/5 + x³/3 - x + arctanx + C

∫x的八次/{x的平方+1}dx
=∫（x^8-1+1)/(x²+1）dx
=∫【（x-1)(x+1)(x²+1）（x^4+1)+1】/(x²+1）dx
=∫（x-1)(x+1)(x^4+1)dx+∫dx/(1+x²）
后面部分就简单啦，∫dx/(1+x²）套公式就是=arctanx
关键就是求∫（x-1)(x+1)(x^4+1)dx
其实也不难
（x-1)(x+1)(x^4+1)=（x²-1）（x^4+1)=x^6-x^4+x²-1
于是
∫（x-1)(x+1)(x^4+1)dx=∫【x^6-x^4+x²-1】dx=x^7/7-x^5/5+x³/3-x
于是∫x的八次/{x的平方+1}dx=x^7/7-x^5/5+x³/3-x+arctanx

令x=tant ，则dx=sec²tdt
原式=∫tant^8/sec²tdx
=∫tant^8/sec²tdtant
=∫tant^8dt
=∫(sec²t-1)^4dt

直接做多项式除法,x^8/(x^2+1)=x^6-x^4+x^2-1+1/(x^2+1)
现在很容易得到结果了.