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求不定积分∫(2x+1)/(x^2+2x-15)dx
人气:499 ℃ 时间:2020-04-08 07:49:04
解答
原式=∫{(2x+2)-1}/(x^2+2x-15)dx=∫(2x+2)/(x^2+2x-15)-1/(x^2+2x-15)dx=ln(x^2+2x-15)-∫1/(x-3)(x+5)dx+c=ln(x^2+2x-15)-1/8∫1/(x-3)-1/(x+5)dx+c=ln(x^2+2x-15)-1/8{ln(x-3)-ln(x+5)}+c
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