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1、高数常见求导数题1.dxx+1((1+3x+1)=.解:令t6=x+1,则dx=6t5dtt=6x+1dxx+1((1+3x+1)=6t5dtt31+t2=6t2dt1+t2=6t2+1-1t2+1dt=61dt-6dtt2+1=6t-6arctant+C∴dxx+1((1+3x+1)=66x+1-6arctan6x+1+C2.dxx2-2x+3=解：dxx2-2x+3=dx2+(x-1)2=1212dx1+(x-12)2=22arctanx-12+C3.dx1+x-x2=.解：dx1+x-x2=dx54-(x2-x+14)=dx54-x-122=dx521-2x-15
2、2=dx521-2x-152=25dx1-2x-152=11-2x-152d2x-15=arcsin2x-15+C4.dx(1+x2)3=解：令x=tant,则dx=1cos2tdt，易知x∈Rt∈(-π2,0)∪(0,π2)，从而有：sint=xcost=x11+tan2t=x1+x2dx(1+x2)3=1cos2tdt(1+tan2t)3=1cos2tdt1cos3t=costdt=sint+C=x1+x2+C∴dx(1+x2)3=x1+x2+C5.X+1X2+X+1dx=解：X+1X2+X+1dx=x+12+12dxx2+x+1=x+12x2+x+1dx+12dx
3、x2+x+1=2x+12x2+x+1dx+12dxx+122+34=x2+x+1+12lnx+12+x+122+34+C常用的积分公式及基本类型(一)1.tanxdx=-lncosx+Ccotxdx=lnsinx+C2.tan2xdx=sec2x-1dx=sec2xdx-dx=tanx-x+C3.cot2xdx=(csc2x-1)dx=cotx-x+C(二)1..sinxcosxdx=sin2x2+C=-cos2x2+C=-cos2x4+C2.cos2xcos3xdx=12cos3x+2x+cos3x-2xdx=12cos5xdx+12cosxdx=110sin5x+1
4、2sinx+Csin3xsin2xdx=12cos3x-2x-cos3x+2xdx=12sinx-110sin5x+C3.cos2xdx=1+cos2x2dx=12dx+14∙2cos2xdx=x2+14sin2x+Csin2xdx=1-cos2x2dx=12dx-142cos2xdx=x2-14sin2x+C4.cos3xdx=cos2x∙cosxdx=(1-sin2x)d(sinx)=sinx-13sin3x+Csin3xdx=sinx2sinxdx=-1-cos2xdcosx=13cos3x-cosx+C(一)1)secxdx=dxcosx=lnsecx+tanx
5、+C证明：令t=secx=1cosx,则x=arccos1t易知x∈0,π2∪π2,π1t∈-1,0∪(0,1];当x∈0,π2secx>0t>0dx=dttt2-1,sec2x-1=tanxsecxdx=t∙dttt2-1=dtt2-1=lnt+t2-1+C=lnsecx+tanx+C当x∈π2,πsecx<0t<0dx=-dttt2-1,sec2x-1=-tanxsecxdx=-t∙dttt2-1=-dtt2-1=-lnt+t2-1+C=lnt-t2-1+C=lnsecx+tanx+C从而有secxdx=dxcosx=lnsecx+tanx+C***sec3xdx=
6、dxcos3x=1cosx1cos2xdx=1cosxdtanx=1cosxtanx-tanxd1cosx=1cosxtanx-sinxcosxsinxcos2xdx=1cosxtanx-1-cos2xcos3xdx=1cosxtanx-dxcos3x+dxcosx=1cosxtanx+lnsecx+tanx-dxcos3dxcos3=12{lnsecx+tanx+1cosxtanx}+C=12lnsecx+tanx+12secxtanx+C1)cscxdx=dxsinx=lncscx-cotx+C证明：令t=cscx=1sinx,则x=arcsin1tdx=-dttt
7、2-1易知x∈-π2,0∪0,π21t∈-1,0∪0,1;当x∈(0,π2]cscx>0t>0dx=-dttt2-1,csc2x-1=cotxsecxdx=t∙-dttt2-1=-dtt2-1=-lnt+t2-1+C=lnt-t2-1+C=ln
8、cscx-cotx
9、+C当x∈-π2,0cscx<0t<0dx=dttt2-1,csc2x-1=-cotxsecxdx=t∙dttt2-1=dtt2-1=lnt+t2-1+C=lncscx-cotx+C从而有cscxdx=dxsinx=lncscx-cotx+C***csc3xdx=dxsin3
