Question

求不定积分，详细步骤

∫xsin(x+1)dx
∫lnx/√x dx
∫e^-x sinx dx
麻烦把答案也写出来还有步骤
第一题和第二题的结果下面的亲好像算的不对哦

Answer 1

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

∫lnx/√x dx
=2∫lnx(1/2√x) dx
=2∫lnxd√x
=2√x*lnx-2∫√xdlnx
=2√x*lnx-2∫√x*1/xd
=2√x*lnx-2∫1/√xd
=2√x*lnx-4∫1/(2√x)d
=2√x*lnx-4√x+C

令a=∫e^-x sinx dx
=-∫e^-xdcosx
=-e^-x cosx+∫cosxde^-x
=-e^-x cosx-∫e^-x cosxdx
=-e^-x cosx-∫e^-x dsinx
=-e^-x cosx-e^-x sinx+∫sinxde^-x
=-e^-x cosx-e^-x sinx-∫e^-x sinxdx
=-e^-x cosx-e^-x sinx-a
所以原式=-1/2e^-x (sinx+cosx)+C

Answer 2

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

∫lnx/√xdx=(1/2)∫lnxd√x
=(1/2)(√x*lnx-∫√xdlnx)
=(1/2)(√x*lnx-∫1/√xdx)
=(1/2)√x*lnx-(1/4)√x+C

∫e^-x sinx dx=-∫sinxde^-x
=∫e^-xdsinx-e^-xsinx=-∫cosxd(e^-x)-(e^-x)sinx=∫e^-xdcosx-(e^-x)cosx-(e^-x)sinx
2∫e^-x sinx dx=-(e^-x)(sin+cosx)
∫e^-x sinx dx=(-1/2)(e^-x)(sin+cosx)+C

Answer 3

1、Sxsin(x+1)dx=-Sxdcos(x+1)=-x*cos(x+1)+Scos(x+1)dx
=-x*cos(x+1)+sin(x+1)+C
2、=-SInxd根号x=-Inx*根号x+S根号xdInx=-INx*根号x+S根号x/xdx
=-Inx*根号x+S1/根号xdx=-Inx*根号x+根号x+C
3、=Ssinxde^-x=sinx*e^-x-Se^-xcosxdx
=sinx*e^-x+Scosxde^-x=sinx*e^-x+cosx*e^-x-Ssinx*e^-xdx
所以2*Se^-xsinxdx=sinx*e^-x+cosx*e^-x
可的结果

Answer 4

第一题，用分部积分法
第二题，令根号x=t，有∫lnx/√x dx=4∫lntdt=……
第三题，也是分布积分