求不定積分解答及過程

應用公式：
∫x^n = (1/n+1)x^(n+1)+ C
∫e^x = e^x + C
∫(1/x)dx = ln|x|


∫(x^-3)(4x^2 +6x +8)dx
= ∫(4x^-1 + 6x^-2 + 8x^-3)dx
= ∫4x^-1dx + ∫6x^-2dx + ∫8x^-3)dx
= 4ln|x| + 6[-x^-1) + 8[(-1/2)x^-2 + C
= 4ln|x| - (6/x) - (4/x^2) + C


∫√x (5x^3 - 6x + 1)dx
= ∫(x)^(1/2)*(5x^3 - 6x + 1)dx
= ∫[5x^(7/5) - 6x^(3/2) + x^(1/2)]dx
= 5(5/12)x^(12/5) - 6(2/5)x^(5/2) + (2/3)x^(3/2) + C
= (25/12)x^(12/5) - (12/5)x^(5/2) + (2/3)x^(3/2)+ C


∫(3/x-2e^x)dx
= ∫(3/x)dx -∫2e^x
= 3ln|x| - 2e^x + C


∫(1/x^2 - 1/3√x + 1/√x)dx
= ∫[x^-2 - (1/3)x^(1/2) + x^(-1/2)]dx
= -x^-1 - (1/3)(2/3)x^(3/2) + (1/2)x^(1/2) + C
= (-1/x)- (2/9)x^(3/2) + (1/2)(√x) + C


∫(2x - 3)^2/x^2 dx
= ∫(4x^2 - 12x + 9)/x^2 dx
= ∫(4 - 12/x + 9x^-2)dx
= 4x - 12ln|x| + 9(-x^-1) + C
= 4x - 12ln|x| - 9/x + C


∫(9x + 6/√x)dx
= ∫[9x + 6x^(-1/2)]dx
= 9(1/2)x^2 + 6[2x^(1/2)] + C
= (9/2)x^2 + 12√x + C

但 ∫[(9x + 6)/√x]dx
= ∫[9x^(1/2) + 6x^(-1/2)]dx
= 9(3/2)x^(3/2) + 6[2x^(1/2)] + C
= (27/2)x^(3/2) + 12√x + C