高數(shù)常見(jiàn)導(dǎo)數(shù)公式推導(dǎo)

1、 高數(shù)常見(jiàn)求導(dǎo)數(shù)題1.d某+1(1+3x+1)=.解:令t6=x+1,則dx=6 t5dtt=6x+1d某+1(1+3x+1)= 6 t5dtt31+t2=6t2dt1+t2=6t2+1-1t2+1dt=61dt-6dtt2+1=6t-6arctant+Cd某+1(1+3x+1)=66x+1-6arctan6x+1+C2.d某2-2x+3=解：d某2-2x+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-152=dx521-2x-152=2

2、5dx1-2x-152=11-2x-152d2x-15=arcsin2x-15+C4.dx(1+x2)3=解：令x=tant,則dx=1cos2tdt，易知xRt(-2,0)(0,2)，從而有：sint=xcost=x11+tan2t=x1+x2dx(1+x2)3=1cos2tdt(1+tan2t)3=1cos2tdt1cos3t=costdt=sint+C=x1+x2+Cdx(1+x2)3=x1+x2+C5.X+1X2+X+1dx=解：X+1X2+X+1dx=x+12+12d某2+x+1=x+12x2+x+1dx+12d某2+x+1=2x+12x2+x+1dx+12d某+122+34=x2+

3、x+1+12lnx+12+x+122+34+C常用的積分公式及基本類型1. tanxdx=-lncosx+Ccotxdx=lnsinx+C2. tan2xdx=sec2x-1dx=sec2xdx-dx=tanx-x+C3. cot2xdx=(csc2x-1)dx=cotx-x+C1. .sinxcosxdx=sin2x2+C=-cos2x2+C=-cos2x4+C2. cos2xcos3xdx=12cos3x+2x+cos3x-2xdx=12cos5xdx+12cosxdx=110sin5x+12sinx+Csin3xsin2xdx=12cos3x-2x-cos3x+2xdx=12sinx-1

4、10sin5x+C3. cos2xdx=1+cos2x2dx=12dx+142cos2xdx=x2+14sin2x+Csin2xdx=1-cos2x2dx=12dx-142cos2xdx=x2-14sin2x+C4. cos3xdx=cos2xcosxdx=(1-sin2x)d(sinx)=sinx-13sin3x+Csin3xdx=sinx2sinxdx=-1-cos2xdcosx=13cos3x-cosx+C1) secxdx=dxcosx=lnsecx+tanx+C證明：令t=secx=1cosx,則x=arccos1t易知x0,22,1t-1,0(0,1;當(dāng)x0,2secx0t0dx=

5、dttt2-1,sec2x-1=tanxsecxdx=tdttt2-1=dtt2-1=lnt+t2-1+C=lnsecx+tanx+C 當(dāng)x2,secx0 t0t0dx=-dttt2-1,csc2x-1=cotxsecxdx=t-dttt2-1=-dtt2-1=-lnt+t2-1+C=lnt-t2-1+C=ln|cscx-cotx|+C當(dāng)x-2,0cscx0t0dx=dttt2-1,csc2x-1=-cotxsecxdx=tdttt2-1=dtt2-1=lnt+t2-1+C=ln cscx-cotx+C從而有cscxdx=dxsinx=lncscx-cotx+C*csc3xdx=dxsin3x

6、=12lncscx-cotx-12cscxcotx+C1) dxa2+x2=1aarctanxa+C2) d某2-a2=12alnx-ax+z+C1) d某2+a2=lnx+x2+a2+C2) d某2-a2=lnx+x2-a2+C3) dxa2-x2=arcsinxa+C1) x2+a2dx=x2x2+a2+a22lnx+x2+a2+C*令x=atant,xRt(-2,2)則dx=acos2tdt，易知有:x2+a2dx= a2tan2t+a2acos2tdt=a21cos3tdt=a22lnsect+tant+a22secttant+C又sect=1cost=tan2t+1x2+a2dx=x

7、2x2+a2+a22lnx2+a2a+xa+C =x2+a2+a22lnx+x2+a2+C從而有x2+a2dx=x2x2+a2+a22lnx+x2+a2+C2) x2-a2dx=x2x2-a2+a22lnx-x2-a2+C*令x=asect,則dx=asintcos2tdt,易知t0,2)(2, 當(dāng)x-a,cost0sint=sec2t-1sec2t=x2-a2-xtant=sectsec2t-1sec2t=-x2-a2a ;x2-a2dx=(asect)2-a2asintcos2tdt=-asintcostasintcos2tdt=a2dtcost-a2dtcos3t=a22lnsect+t

8、ant-a22secttant=x2x2-a2+a22lnx-x2-a2+C 當(dāng)xa,cost0,sint0sint=sec2t-1sec2t=x2-a2xtant=sectsec2t-1sec2t=x2-a2a;x2-a2dx=asect2-a2asintcos2tdt=asintcostasintcos2tdt=a2dxcos3t-a2dxcost=-a22lnsect+tant+a22secttant=x2x2-a2+a22lnx-x2-a2+C從而有x2-a2dx=x2x2-a2+a22lnx-x2-a2+C3)a2-x2dx=x2a2-x2-a22arccosxa+C*令x=acost,則dx=-asintdt ,易知有t0,sint0顯然可知，不管x0或x0,都有下列式子成立a2-x2dx=a2-a2cos2t-asintdt=-a2sin2tdt=a2cos