Question

不定积分，求原函数

求1/（1+x*x*x*x）的原函数

Answer 1

答案在图片上

Answer 2

解：1+x^4 = (1+x²)² - 2x² = (1+x²+√2x)(1+x²-√2x)
1/(1+x^4)
= [1/(1+x²-√2x) - 1/(1+x²+√2x)]/2√2x
= 1/2√2 *[1/x + (√2-x)/(1+x²-√2x) - 1/x + (√2+x)/(1+x²+√2x)]
= 1/4√2 * [(2x+2√2)/(x²+√2x+1) - (2x-2√2)/(x²+1-√2x)]
= 1/4√2 *[(2x+√2)/(x²+√2x+1) - (2x-√2)/(x²+1-√2x) + √2/(x²+√2x+1) + √2/(x²+1-√2x)]
对(2x+√2)/(x²+√2x+1)求积分得ln(x²+√2x+1)
对(2x-√2)/(x²+1-√2x)求积分得ln(x²+1-√2x)
对√2/(x²+√2x+1)求积分得2arctan(√2x+1)
对√2/(x²-√2x+1)求积分得2arctan(√2x-1)
原式 = 1/4√2 *{ln[(x²+√2x+1))/(x²+1-√2x)] + 2arctan(√2x+1) + 2arctan(√2x-1)} + C