作业帮求arctanx/(x2(1+x2))的不定积分?
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求arctanx/(x2(1+x2))的不定积分?
求arctanx/(x2(1+x2))的不定积分?
求arctanx/(x2(1+x2))的不定积分?
∫arctanxdx/[x^2(1+x^2)]
=∫arctanxdx/x^2 -∫arctanxdx/(1+x^2)
=∫arctanxd(-1/x)-∫arctanxdarctanx
=-(arctanx)/x +∫(1/x)darctanx-(arctanx)^2/2
=-(arctanx)/x-(arctanx)^2/2+∫dx/[x(1+x^2)]
其中 ∫dx/[x(1+x^2)]=∫[(1+x^2)-x^2]dx/[x(1+x^2)]=∫dx/x-∫xdx/(1+x^2)=lnx-(1/2)ln(1+x^2)+C
原式=-(arctanx)/x-(arctanx)^2/2+lnx-(1/2)ln(1+x^2)+C
1/(x2(1+x2))=1/x^2-1/(x^2+1)
1/x *1/(x^2+1)=1/x-x/(x^2+1)
S[arctanx/(x2(1+x2))]dx
=S[arctanx*(1/x^2-1/(x^2+1))dx
=Sarctanx*1/x^2dx-Sarctanx*1/(x^2+1)dx
=-Sarctanx d(1/x)-Sarctanx d...
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1/(x2(1+x2))=1/x^2-1/(x^2+1)
1/x *1/(x^2+1)=1/x-x/(x^2+1)
S[arctanx/(x2(1+x2))]dx
=S[arctanx*(1/x^2-1/(x^2+1))dx
=Sarctanx*1/x^2dx-Sarctanx*1/(x^2+1)dx
=-Sarctanx d(1/x)-Sarctanx darctanx
=-arctanx *1/x+S1/x *darctanx -(arctanx )^2 *1/2
=-arctanx *1/x-(arctanx )^2 *1/2+S (1/x *1/(x^2+1)*dx
=-arctanx *1/x-(arctanx )^2 *1/2+S(1/x)*dx-S(x/(x^2+1)dx
=-arctanx *1/x-(arctanx )^2 *1/2+lnx-1/2*S1/(x^2+1) d(x^2+1)
==-arctanx *1/x-(arctanx )^2 *1/2+lnx-1/2*ln(x^2+1)+c
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