∫(-1,1)x/√(5-4x)dx
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∫[(x³-17)/(x²-4x+3)]dx
=∫[(x³-4x²+3x+4x²-16x+12+13x-29)/(x²-4x+3)]dx
=∫[x(x²-4x+3)+4(x²-4x+3)+(5x-5+8x-24)]/(x²-4x+3)]dx
=∫[x+4 + (5x-5+8x-24)/(x-1)(x-3)]dx
=∫[x+4 + 5/(x-3) +8/(x-1)]dx
=∫(x+4)dx+ 5∫[1/(x-3)]d(x-3) +8∫[1/(x-1)]d(x-1)
=½x²+4x+ 5ln|x-3|+8ln|x-1|+C
=∫[(x³-4x²+3x+4x²-16x+12+13x-29)/(x²-4x+3)]dx
=∫[x(x²-4x+3)+4(x²-4x+3)+(5x-5+8x-24)]/(x²-4x+3)]dx
=∫[x+4 + (5x-5+8x-24)/(x-1)(x-3)]dx
=∫[x+4 + 5/(x-3) +8/(x-1)]dx
=∫(x+4)dx+ 5∫[1/(x-3)]d(x-3) +8∫[1/(x-1)]d(x-1)
=½x²+4x+ 5ln|x-3|+8ln|x-1|+C
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