首先,可以将原函数表示为 f(x) = 3(x-4)^2(x+1)^8,然后进行变量代换 u = x-4,得到:
f(u+4) = 3(u)^2(u+5)^8
接着,将 u+4 替换回 x,得到:
f(x) = 3(x-4)^2(x+1)^8 = 3(u)^2(u+5)^8 = 3(x-4)^2(x-3)^8
现在可以对 f(x) 进行积分,得到:
∫f(x)dx = ∫3(x-4)^2(x+1)^8dx
= ∫3(u)^2(u+5)^8du (使用变量代换)
= ∫3x^2(x-3)^8dx (将 u 替换回 x)
= ∫3(x^2-6x+9)(x^8-24x^7+252x^6-1512x^5+5670x^4-14112x^3+21870x^2-19656x+7296)dx
展开后,可以得到一些多项式的积分,利用多项式积分的公式,可以求出:
∫x^2(x-3)^8dx = (1/9)*x^9 - (3/4)*x^8 + (27/7)*x^7 - 27x^6 + (324/5)*x^5 - (3024/5)*x^4 + (2187/2)*x^3 - (13122/7)*x^2 + (19683/10)*x
∫x^3(x-3)^8dx = (1/10)*x^10 - (3/5)*x^9 + (27/4)*x^8 - (27/2)*x^7 + (324/7)*x^6 - (504/5)*x^5 + (6561/14)*x^4 - (39366/35)*x^3 + (19683/56)*x^2
将上述结果代回原式,得到:
∫f(x)dx = 3*[(1/9)*x^9 - (3/4)*x^8 + (27/7)*x^7 - 27x^6 + (324/5)*x^5 - (3024/5)*x^4 + (2187/2)*x^3 - (13122/7)*x^2 + (19683/10)x] - 3[(1/10)*x^10 - (3/5)*x^9 + (27/4)*x^8 - (27/2)*x^7 + (324/7)*x^6 - (504/5)*x^5 + (6561/14)*x^4 - (39366/35)*x^3 + (19683/56)*x^2] + C
其中 C 是常数项,可以根据初始条件计算。
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