(1)设2t=x-3，则dx=2dt，
∴∫[(x+5)/(x²-6ⅹ+13)]dx
=∫[(2t+8)/(4t²+4)]·2dt
=(1/2)∫[1/(t²+1)]d(t²+1)+4∫[1/(t²+1)]dt
=(1/2)㏑|t²+1|+4arctant+C
=(1/2)㏑|[(x-3)/2]²+1|+4arctan[(x-3)/2]+C
(2)设x²+1=2t，则ⅹdx=tdt
∴∫[x/(x^4+2x²+5)]dx
=(1/4)∫[1/(t²+1)]dt
=(1/4)arctant+C
=(1/4)arctan[(x²+1)/2]+C
(3)∫[x³/√(1+x²)]dx
=∫[(1+x²-1)/√(1+x²)]d(1+x²)
=∫√(1+x²)d(1+x²)-∫[1/√(1+x²)]d(1+x²)
=(2/3)√(1+x²)³-2√(1+x²)+C
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