求定积分∫(上限8,下限0)dx/1+3√x

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a=3√xx=a²/9
dx=2ada/9
x=8,a=6√2
x=0,a=0
原式=∫(上限6√2,下限0)(2ada/9)/(1+a)
=(2/9)∫(上限6√2,下限0)ada/(1+a)
=(2/9)∫(上限6√2,下限0)(a+1-1)da/(1+a)
=(2/9)∫(上限6√2,下限0)[1-1/(a+1)]da
=(2/9)[a-ln(a+1)](上限6√2,下限0)
=(2/9)[6√2-ln(6√2+1)-0+0]
=4√2/3-2ln(6√2+1)/9
======以下答案可供参考======
供参考答案1:
上位仁兄结果是对地
答案也是正确地~~