急需答案,跪求
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(1)分部积分法
∫xln(x-1)dx=1/2∫ln(x-1)dx²
=x²lnx/2-1/2∫x²dln(x-1)
=x²lnx/2-1/2∫x²/(x-1)dx
=x²lnx/2-1/2∫x+1+1/(x-1)dx
=x²lnx/2-x²/4-x/2-ln(x-1)/2+C
(2)令a=√x
x=a²
dx=2ada
原式=∫cosa*2ada
=2∫adsina
=2asina-2∫sinada
=2asina+2cosa+C
=2√xsin√x+2cos√x+C
∫xln(x-1)dx=1/2∫ln(x-1)dx²
=x²lnx/2-1/2∫x²dln(x-1)
=x²lnx/2-1/2∫x²/(x-1)dx
=x²lnx/2-1/2∫x+1+1/(x-1)dx
=x²lnx/2-x²/4-x/2-ln(x-1)/2+C
(2)令a=√x
x=a²
dx=2ada
原式=∫cosa*2ada
=2∫adsina
=2asina-2∫sinada
=2asina+2cosa+C
=2√xsin√x+2cos√x+C
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