∫[0→1] 1/(1+2x)³ dx
=(1/2)∫[0→1] 1/(1+2x)³ d(2x)
=(1/2)(-1/2)[1/(1+2x)²] |[0→1]
=-1/[4(1+2x)²] |[0→1]
=2/9
2、∫[1→2] √(x²-1)/x dx
先做不定积分
令x=secu，则√(x²-1)=tanu，dx=secutanudu
∫ √(x²-1)/x dx
=∫ (tanu/secu)(secutanu) du
=∫ tan²u du
=∫ (sec²u - 1) du
=tanu - u + C
=√(x²-1) - arccos(1/x) + C
代入上下限相减得：√3 - π/3
