求解一道数学定积分题,想知道怎么求得的,在线坐等大神解答 一道定积分练习题 求解答!
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现在我以e为底来作。
求定积分[0,1]∫[4^(2x)]ln(4^x+8)dx
解:令4^x+8=y,则4^x=y-8,于是d(4^x)=(4^x)ln4dx=dy,故dx=dy/[(4^x)ln4]=dy/[(y-8)ln4];
4^(2x)=(4^x)²=(y-8)²;x=0时y=9;x=1时y=12;代入原式得:
原式=[9,12]∫(y-8)²(lny)/[(y-8)ln4]dy=[9,12](1/ln4)∫(y-8)(lny)dy=[9,12](1/ln4)[∫ylnydy-8∫lnydy]
=[9,12](1/ln4)[(1/2)∫lnyd(y²)-8(ylny-∫dy)]=[9,12](1/ln4){(1/2)[y²lny-∫ydy]-8(ylny-y)}
=(1/ln4)[(1/2)(y²lny-y²/2)-8(ylny-y)]∣[9,12]
=(1/ln4){[(1/2)lny-(1/4)]y²-8y(lny-1)}∣[9,12]
=(1/ln4){[(1/2)ln12-(1/4)]×144-96(ln12-1)-[(1/2)ln9-(1/4)]×9+72(ln9-1)}
=(-177ln3-2ln2+537/4)/ln4.
这是一个简单的换元法积分
过程稍等
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