∫2x(x2+1)(x2+3)dx
=2∫x(x2+1)(x2+3)dx
=2∫x2(x2+1)−x2(x2+3)dx
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∫x2(x2+1)
substitute u=x2+1→du=2xdx
=12∫12udu
=14ln|x2+1|
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∫x2(x2+3)
substitute u=x2+3→du=2xdx
=12∫12udu
=14ln|x2+3|
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=2(14ln|x2+1|−14ln|x2+3|)+C