求此题有理不定积分证明
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书上这个递推公式有误
A(k)=∫dt/(t^2+r^2)^k
=t/(t^2+r^2)^k-∫td[1/(t^2+r^2)^k]
=t/(t^2+r^2)^k+2k∫t^2/(t^2+r^2)^(k+1)dt
=t/(t^2+r^2)^k+2k∫[1/(t^2+r^2)^k-r^2/(t^2+r^2)^(k+1)]dt
=t/(t^2+r^2)^k+2k*A(k)-2kr^2*A(k+1)
则A(k+1)=[1/(2kr^2)]*[t/(t^2+r^2)^k+(2k-1)*A(k)]
所以A(k)=[1/2(k-1)r^2]*[t/(t^2+r^2)^(k-1)+(2k-3)*A(k-1)]
A(k)=∫dt/(t^2+r^2)^k
=t/(t^2+r^2)^k-∫td[1/(t^2+r^2)^k]
=t/(t^2+r^2)^k+2k∫t^2/(t^2+r^2)^(k+1)dt
=t/(t^2+r^2)^k+2k∫[1/(t^2+r^2)^k-r^2/(t^2+r^2)^(k+1)]dt
=t/(t^2+r^2)^k+2k*A(k)-2kr^2*A(k+1)
则A(k+1)=[1/(2kr^2)]*[t/(t^2+r^2)^k+(2k-1)*A(k)]
所以A(k)=[1/2(k-1)r^2]*[t/(t^2+r^2)^(k-1)+(2k-3)*A(k-1)]
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