Question

不定积分 1/x^4√(1+x^2) dx

Answer 1

∫ 1/[x^4 * √(1+x²)] dx
Let x = tany and dx = sec²y dy,then √(1+x²) = secy
==> ∫ sec²y / (tan^4y * secy) dy
= ∫ secy * cos^4y / sin^4y dy
= ∫ cscy * cot³y dy
= ∫ cscy * coty * (csc²y - 1) dy
= -∫ (csc²y - 1) dcscy
= cscy - csc³y / 3 + C
= [(2x²-1)√(1+x²)] / (3x³) + C

楼上的答案不对，-1/(3x³) + 1/x - arctan(1/x)的导数是1/(x^4 + x^6) = 1/[x^4 * (1+x²)]
漏了个根号

Answer 2

倒代换的题目x=1/t,dx=-1/t^2
∫(-1/t^2)dt/[1/t^4(1+1/t^2)]
=-∫t^4dt/(t^2+1)
=-∫(t^4-1+1)/(t^2+1)
=-∫(t^2-1)dt-∫1/(t^2+1) dt
=-t^3/3+t-arctant+C

答案是-1/(3*x^3)+1/x-arctan(1/x)+C