Question

一道定积分题 求解答过程

Answer 1

解：
∫(x^4)/(1-x^4)dx
=-∫[(x^4)-1+1]/[(x^4)-1]dx
=-∫{1+1/[(x^4)-1]}dx
=-∫1dx-∫dx/[(x^4)-1]
=-x-∫dx/[(x^4)-1]
=-x-∫dx/[(x²+1)(x²-1)]
=-x-(1/2)∫[1/(x²-1)-1/(x²+1)]dx
=-x-(1/2)∫{1/[(x+1)(x-1)]-1/(x²+1)}dx
=-x-(1/2)∫dx/[(x+1)(x-1)]+(1/2)∫dx/(x²+1)
=-x-(1/4)∫[1/(x-1)-1/(x+1)]dx+(1/2)∫dx/(x²+1)
=-x-(1/4)∫dx/(x-1)+(1/4)∫dx/(x+1)+(1/2)∫dx/(x²+1)
=-x-[ln(x-1)]/4+[ln(x+1)]/4+(1/2)∫dx/(x²+1)
=-x-(1/4)ln[(x-1)/(x+1)]+(1/2)∫dx/(x²+1)………………(1)
对于∫dx/(x²+1)……………………………………………(2)
设：x=tany，则y=arctanx、dx=d(tany)=dy/cos²y、1/(x²+1)=cos²y
代入(2)，有：
∫dx/(x²+1)
=∫(cos²y/cos²y)dy
=y+C
=arctanx+C
代入(1)，有：
∫(x^4)/(1-x^4)dx=-x-(1/4)ln[(x-1)/(x+1)]+(arctanx)/2+C

Answer 2