∫dx/(³√x+⁴√x)化简
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∫dx/(³√x+⁴√x)化简
![∫dx/(³√x+⁴√x)化简](/uploads/image/z/19966265-17-5.jpg?t=%E2%88%ABdx%2F%EF%BC%88%26%23179%3B%E2%88%9Ax%2B%26%238308%3B%E2%88%9Ax%EF%BC%89%E5%8C%96%E7%AE%80)
1/3和1/4的公倍数是1/2,所以令t = x^(12),x = t^12,dx = 12t^11 dt
∫ 1/[x^(1/3) + x^(1/4)] dx
= ∫ 1/(t^4 + t^3) * 12t^11 dt
= 12∫ t^8/(1 + t) dt,不断用a^2 - b^2 = (a - b)(a + b)化简
= 12∫ [(t^8 - 1) + 1]/(1 + t) dt
= 12∫ [(t^4 - 1)(t^4 + 1) + 1]/(1 + t) dt
= 12∫ [(t^2 - 1)(t^2 + 1)(t^4 + 1) + 1]/(1 + t) dt
= 12∫ [(t - 1)(t + 1)(t^2 + 1)(t^4 + 1) + 1]/(1 + t) dt
= 12∫ (t - 1)(t^2 + 1)(t^4 + 1) dt + 12∫ dt/(1 + t)
= 12∫ (t^7 - t^6 + t^5 - t^4 + t^3 - t^2 + t - 1) dt + 12ln|1 + t| + C
= 12[(1/8)t^8 - (1/7)t^7 + (1/6)t^6 - (1/5)t^5 + (1/4)t^4 - (1/3)t^3 + (1/2)t^2 - t] + 12ln|1 + t| + C
= (3/2)t^8 - (12/7)t^7 + 2t^6 - (12/5)t^5 + 3t^4 - 4t^3 + 6t^2 - 12t + 12ln|1 + t| + C
= (3/2)x^(2/3) - (12/7)x^(7/12) - (12/5)x^(5/12) + 2x^(1/2) + 3x^(1/3) - 4x^(1/4) + 6x^(1/6) - 12x^(1/12) + 12ln[1 + x^(1/2)] + C
∫ 1/[x^(1/3) + x^(1/4)] dx
= ∫ 1/(t^4 + t^3) * 12t^11 dt
= 12∫ t^8/(1 + t) dt,不断用a^2 - b^2 = (a - b)(a + b)化简
= 12∫ [(t^8 - 1) + 1]/(1 + t) dt
= 12∫ [(t^4 - 1)(t^4 + 1) + 1]/(1 + t) dt
= 12∫ [(t^2 - 1)(t^2 + 1)(t^4 + 1) + 1]/(1 + t) dt
= 12∫ [(t - 1)(t + 1)(t^2 + 1)(t^4 + 1) + 1]/(1 + t) dt
= 12∫ (t - 1)(t^2 + 1)(t^4 + 1) dt + 12∫ dt/(1 + t)
= 12∫ (t^7 - t^6 + t^5 - t^4 + t^3 - t^2 + t - 1) dt + 12ln|1 + t| + C
= 12[(1/8)t^8 - (1/7)t^7 + (1/6)t^6 - (1/5)t^5 + (1/4)t^4 - (1/3)t^3 + (1/2)t^2 - t] + 12ln|1 + t| + C
= (3/2)t^8 - (12/7)t^7 + 2t^6 - (12/5)t^5 + 3t^4 - 4t^3 + 6t^2 - 12t + 12ln|1 + t| + C
= (3/2)x^(2/3) - (12/7)x^(7/12) - (12/5)x^(5/12) + 2x^(1/2) + 3x^(1/3) - 4x^(1/4) + 6x^(1/6) - 12x^(1/12) + 12ln[1 + x^(1/2)] + C