用分部积分法求下列不定积分 1）∫xsin2xdx 2）∫xlnxdx 3）∫arccosxdx 4）∫xarctanxdx

Question

用分部积分法求下列不定积分

1）∫xsin2xdx
2）∫xlnxdx
3）∫arccosxdx
4）∫xarctanxdx

Answer 1

2）3）4）答案同楼上，
1）∫xsin2xdx=(-1/2)∫xdcos2x=(-1/2)xcos2x+(1/2)∫cos2xdx=(-1/2)xcos2x+(1/4)sin2x+C

2）∫xlnxdx=(1/2)∫lnxdx^2=(1/2)x^2lnx-(1/2)∫xdx=(1/2)x^2lnx-(1/4)x^2+C

3）∫arccosxdx=xarccosx-∫-xdx/√(1-x^2)
=xarctanx-(1/2)d(1-x^2)/√(1-x^2)
=xarccosx -√(1-x^2)+C

4）∫xarctanxdx=(1/2)∫arctanxdx^2 =(1/2)x^2arctanx-(1/2)∫x^2dx/(1+x^2)
=(1/2)x^2arctanx-(1/2)x+(1/2)∫dx/(1+x^2)
=(1/2)x^2arctnax-(x/2)+(1/2)arctanx+C

Answer 2

1）∫xsin2xdx
=S1/2*xdsin2x
=1/2*xsin2x-1/2*Ssin2xdx
=1/2*xsin2x-1/4*Ssin2xd2x
=1/2*xsin2x+1/4*cos2x+c
2）∫xlnxdx
=1/2*Slnxdx^2
=1/2*x^2*lnx-1/2*Sx^2dlnx
=1/2*x^2 *lnx-1/2*Sx^2*1/x*dx
=1/2*x^2*lnx-1/2*Sxdx
=1/2*x^2*lnx-1/4*x^2+c

3）∫arccosxdx
=xarccosx-Sxdarccosx
=xarccosx-Sx*(-1/(1-x^2)^(1/2))dx
=xarccosx+Sx/根号（1-x^2)dx
=xarccosx-根号（1-x^2)+c
4）∫xarctanxdx
=1/2*Sarctanxdx^2
=1/2*x^2*arctanx-1/2*Sx^2darctanx
=1/2*x^2*arctanx-1/2*Sx^2/(x^2+1)dx
=1/2*x^2*arctanx-1/2*Sdx+1/2*S1/(x^2+1)dx
=1/2*x^2*arctanx-x/2+1/2*arctanx+c

Answer 3

1. 原式=x^2 /2 * sin2x - 1/2 ∫sin2xd2x = x^2 /2 * sin2x - 1/2 * cos2x + c
2. 原式=x^2 /2 * lnx - ∫lnxdx = x^2 /2 * lnx -1/x + c
3. 原式=x * arccosx + ∫x/根号下(1-x^2) , 令x=sint，得 x * arccosx + 根号下(1-x^2)