F(6)
∫log<5>(2x-1) dx
=(1/ln5)∫ln(2x-1) dx
=(1/ln5)x.ln(2x-1)-(1/ln5)∫ 2x/(2x-1) dx
=(1/ln5)x.ln(2x-1)-(1/ln5)∫[ 1+ 1/(2x-1) ]dx
=(1/ln5)x.ln(2x-1)-(1/ln5)[ x+ (1/2)ln|2x-1| ] +C
(1)
lim(x->0) ( 1-1/x)^(2x)
=e^(-2)
F(9)
∫x^2.e^x dx
=∫x^2 de^x
=x^2.e^x -2∫xe^x dx
=x^2.e^x -2∫x de^x
=x^2.e^x -2xe^x +2∫e^x dx
=x^2.e^x -2xe^x +2e^x +C
F(10)
∫(log<2>x)^2 dx
=(1/ln2)^2.∫(lnx)^2 dx
=(1/ln2)^2. x(lnx)^2 -[2/(ln2)^2]∫ lnx dx
=(1/ln2)^2. x(lnx)^2 -[2/(ln2)^2]xlnx +[2/(ln2)^2]∫ dx
=(1/ln2)^2. x(lnx)^2 -[2/(ln2)^2]xlnx +[2/(ln2)^2]x + C
F(13)
∫sin(lnx)dx
=xsin(lnx) -∫cos(lnx)dx
=xsin(lnx) -xcos(lnx) -∫sin(lnx)dx
2∫sin(lnx)dx =xsin(lnx) -xcos(lnx)
∫sin(lnx)dx =(1/2)[xsin(lnx) -xcos(lnx)] + C
F(14)
∫x.arcsinx/√(1-x^2) dx
=-∫arcsinx d√(1-x^2)
=-arcsinx.√(1-x^2) +∫dx
=-arcsinx.√(1-x^2) +x + C
F(15)
∫ln(lnx)/x dx
=∫ln(lnx) dlnx
=(1/2)[ln(lnx)]^2 +C
追问十四小题的第一个等号怎么得到
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/838ba61ea8d3fd1f8122735c3f4e251f95ca5fa9?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
十五题这样写哪错了了呢
追答F(15)
∫ln(lnx)/x dx
=∫ln(lnx) dlnx
=lnx.ln(lnx) -∫ (1/x) dx
=lnx.ln(lnx) -ln|x| + C
追问和你一开始做的答案是一样的吗
追答不!