用第一类换元积分法求下列不定积分

(1)令u = 4x,du = 4dx
∫ cos4x dx
= (1/4)∫ cosu du
= (1/4)sinu + C
= (1/4)sin(4x) + C
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(2)令u = - 2x,du = - 2 dx
∫ e^(- 2x) dx
= (- 1/2)∫ e^u du
= (- 1/2)e^u + C
= (- 1/2)e^(- 2x) + C
= - 1/[2e^(2x)] + C
————————————————————————————
(3)令u = 3x - 1,du = 3 dx
∫ (3x - 1)^4 dx
= (1/3)∫ u^4 du
= (1/3)(1/5)u^5 + C
= (1/15)(3x - 1)^5 + C
————————————————————————————
(4)令u = 3x,du = 3 dx
∫ 10^(3x) dx
= (1/3)∫ 10^u du
= (1/3) * (10^u)/ln(10) + C
= (10^u)/[3ln(10)] + C
= [10^(3x)]/[3ln(10)] + C

⑴∫cos4x dx=(1/4)∫cos4x d4x=(1/4)sin4x+C
⑵∫e^-2x dx=(-1/2)∫e^-2x d(-2x)=(-1/2)e^(-2x)+C
⑶∫ (3x－1)^4 dx=（1/3）∫ (3x－1)^4 d（3x-1）=（1/3）（1/5)(3x-1)^5=(1/15)（3x-1)^5 +C
⑷∫10^3x dx=(1/3∫10^3x d3x=(1/3)10^(3X)/ln10+C=10^(3x)/3ln10 +C