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∫xCos(x+1)Dx的不定积分

=∫xdsin(x+1) =xsin(x+1)-∫sin(x+1)dx =xsin(x+1)+cos(x+1)+C

∫xln(x²+1)dx=(1/2)∫ln(x²+1)d(x²+1)=(1/2)∫lntdt (t=x²+1) 然后用分部积分法,结果是(1/2)(x²+1)ln(x²+1)-(x²+1)+c

t=√(x+1),x=t^2-1,dx=2tdt 原积分=S(t^2-1)/t 2tdt=S2(t^2-1)dt=2/3*t^3-t+c=2/3(x+1)^(3/2)-2(x+1)^(1/2)+c

∫1/x(x²+1)dx =∫1/x-x/(x²+1)dx =∫1/xdx-∫x/(x²+1)dx =ln|x|-1/2ln|x²+1|+c

如图

答案如图所示

【xlnx】′=1+lnx 所以对lnx积分=xlnx -x 【x²lnx】=2xlnx+x所以对2xlnx积分=x²lnx-x²/2 ∫xln(x-1)dx =∫【(x-1)ln(x-1)+ln(x-1)】d(x-1) 分别积分 =0.5*(x-1)²ln(x-1)-0.25(x-1)² + (x-1)ln(x-1)-(x-1)+C...

x=t² ∫x/(1+√x)dx=∫t²/(1+t)dt² =2∫t³/(1+t)dt=2∫(t³+t²-t²-t+t+1-1)/(1+t)dt = 2∫t²-t+1-1/(1+t)dt =2(t³/3-t²/2+t-ln|1+t)+C t=√x ....

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