高数求救用分部积分法求∫上限1下限0 （x-1）3^xdx.麻烦了~
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∫[0→1] (x - 1)3^x dx
= ∫[0→1] (x - 1) d[3^x/ln3]
= (1/ln3)(x - 1)3^x |[0→1] - (1/ln3)∫[0→1] 3^x d(x - 1)
= (1/ln3)[0 - (0 - 1)(1)] - (1/ln3)∫[0→1] 3^x dx
= 1/ln3 - 1/ln3 * [3^x/ln3] |[0→1]
= 1/ln3 - 1/(ln3)^2 * [3 - 1]
= (ln3 - 2)/(ln3)^2 ≈ - 0.7468
= ∫[0→1] (x - 1) d[3^x/ln3]
= (1/ln3)(x - 1)3^x |[0→1] - (1/ln3)∫[0→1] 3^x d(x - 1)
= (1/ln3)[0 - (0 - 1)(1)] - (1/ln3)∫[0→1] 3^x dx
= 1/ln3 - 1/ln3 * [3^x/ln3] |[0→1]
= 1/ln3 - 1/(ln3)^2 * [3 - 1]
= (ln3 - 2)/(ln3)^2 ≈ - 0.7468
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