大学数学。不定积分?
3个回答
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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
∫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
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