高數(shù)常見導(dǎo)數(shù)公式推導(dǎo)(已修改)
2025-07-08 20:48 本頁面
【正文】 .. . . ..高數(shù)常見求導(dǎo)數(shù)題+1((1+3x+1) =. 解:令t6=x+1,則dx=6 t5dtt=6x+1 dxx+1((1+3x+1) = 6 t5dtt31+t2= 6t2dt1+t2=6t2+11t2+1dt=61dt6dtt2+1=6t6arctant+C∴dxx+1((1+3x+1)=66x+16arctan6x+1+C2. dxx22x+3 = 解：dxx22x+3 =dx2+(x1)2=1212dx1+(x12)2=22arctanx12+C+xx2= .解：dx1+xx2=dx54(x2x+14)=dx54x122= dx5212x152=dx5212x152=25dx12x152=112x152d2x15=arcsin2x15+C4. dx(1+x2)3= 解：令x=tant,則dx=1 cos2tdt，易知x∈Rt∈(π2,0)∪(0,π2)，從而有：sint=xcost=x11+tan2t=x1+x2dx(1+x2)3=1 cos2tdt(1+tan2t)3= 1 cos2tdt1cos3t=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+12dxx2+
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