求不定积分∫2x^3(x^2+1)^{1/2}dx
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∫[1/(x²-2x-3)]dx
=∫[1/(x+1)(x-3)]dx
=¼∫[(x+1)-(x-3)]/[(x+1)(x-3)] dx
=¼∫[1/(x-3) -1/(x+1)]dx
=¼∫[1/(x-3)]d(x-3) -¼∫[1/(x+1)]d(x+1)
=¼ln|x-3|-¼|ln(x+1)|+C
=¼ln|(x-3)/(x+1)| +C
=∫[1/(x+1)(x-3)]dx
=¼∫[(x+1)-(x-3)]/[(x+1)(x-3)] dx
=¼∫[1/(x-3) -1/(x+1)]dx
=¼∫[1/(x-3)]d(x-3) -¼∫[1/(x+1)]d(x+1)
=¼ln|x-3|-¼|ln(x+1)|+C
=¼ln|(x-3)/(x+1)| +C
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