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æ±å®ç§¯å[0ï¼1]â«[4^(2x)]ln(4^x+8)dx
解ï¼ä»¤4^x+8=yï¼å4^x=y-8ï¼äºæ¯d(4^x)=(4^x)ln4dx=dyï¼æ
dx=dy/[(4^x)ln4]=dy/[(y-8)ln4]ï¼
4^(2x)=(4^x)²=(y-8)²ï¼x=0æ¶y=9ï¼x=1æ¶y=12ï¼ä»£å
¥åå¼å¾ï¼
åå¼=[9ï¼12]â«(y-8)²(lny)/[(y-8)ln4]dy=[9ï¼12](1/ln4)â«(y-8)(lny)dy=[9ï¼12](1/ln4)[â«ylnydy-8â«lnydy]
=[9ï¼12](1/ln4)[(1/2)â«lnyd(y²)-8(ylny-â«dy)]=[9ï¼12](1/ln4){(1/2)[y²lny-â«ydy]-8(ylny-y)}
=(1/ln4)[(1/2)(y²lny-y²/2)-8(ylny-y)]â£[9ï¼12]
=(1/ln4){[(1/2)lny-(1/4)]y²-8y(lny-1)}â£[9ï¼12]
=(1/ln4){[(1/2)ln12-(1/4)]Ã144-96(ln12-1)-[(1/2)ln9-(1/4)]Ã9+72(ln9-1)}
=(-177ln3-2ln2+537/4)/ln4.
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