求∫(x+1)/(x^2-5x+6)dx
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原式=∫(x+1)/[(x-2)(x+3)]dx
=∫(x+3-2)/[(x-2)(x+3)]dx
=∫(x+3)/[(x-2)(x+3)]dx-∫2/[(x-2)(x+3)]dx
=∫dx/(x-2)-2∫1/[(x-2)(x+3)]dx
=ln(x-2)-2/5∫[1/(x-2)-1/(x+3)]dx
=ln(x-2)-2/5[∫1/(x-2)dx-∫1/(x+3)dx]
=ln(x-2)-2/5[ln(x-2)-ln(x+3)]
=3/5ln(x-2)+2/5ln(x+3)
有不明白之处请留言.
=∫(x+3-2)/[(x-2)(x+3)]dx
=∫(x+3)/[(x-2)(x+3)]dx-∫2/[(x-2)(x+3)]dx
=∫dx/(x-2)-2∫1/[(x-2)(x+3)]dx
=ln(x-2)-2/5∫[1/(x-2)-1/(x+3)]dx
=ln(x-2)-2/5[∫1/(x-2)dx-∫1/(x+3)dx]
=ln(x-2)-2/5[ln(x-2)-ln(x+3)]
=3/5ln(x-2)+2/5ln(x+3)
有不明白之处请留言.
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