1,问题转化为求∫x/(x+1)^4dx和∫1/(x+1)^4dx
只需求∫x/(x+1)^4dx=∫dx/(x+1)^2-∫xdx/(x+1)^3-∫1/(x+1)^4dx
只需求∫xdx/(x+1)^3=∫dx/(x+1)-∫xdx/(x+1)^2-∫dx/(x+1)^3
只需求∫xdx/(x+1)^2=∫dx/(x+1)-∫dx/(x+1)^2
问题解决.
2注意到d[1/(xcosx-sinx)]=xsinxdx/(xcosx-sinx)^2
原式=∫(sinx/x)*d[1/(xcosx-sinx)]
=(sinx/x)*1/(xcosx-sinx)-∫[(xcosx-sinx)/x^2]*[1/(xcosx-sinx)]dx
=(sinx/x)*1/(xcosx-sinx)-∫dx/x^2
=(sinx/x)*1/(xcosx-sinx)+1/x+C