∫1/(x(√x+x^(2/5)))dx 详细过程！！谢谢
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解：设x=t^10，则dx=10t^9*dt
∴原式=∫10t^9*dt/[t^10(t^5+t^4)]
=∫[1/t-1/t²+1/t³-1/t^4+1/t^5-1/(t+1)]dt
=ln│t│+1/t-(1/2)/t²+(1/3)/t³-(1/4)/t^4-ln│t+1│+C (C是积分常数)
=ln│t/(t+1)│+1/t-(1/2)/t²+(1/3)/t³-(1/4)/t^4+C
=ln│x^(1/10)/(x^(1/10)+1)│+1/x^(1/10)-(1/2)/x^(1/5)+(1/3)/x^(3/10)-(1/4)/x^(2/5)+C
∴原式=∫10t^9*dt/[t^10(t^5+t^4)]
=∫[1/t-1/t²+1/t³-1/t^4+1/t^5-1/(t+1)]dt
=ln│t│+1/t-(1/2)/t²+(1/3)/t³-(1/4)/t^4-ln│t+1│+C (C是积分常数)
=ln│t/(t+1)│+1/t-(1/2)/t²+(1/3)/t³-(1/4)/t^4+C
=ln│x^(1/10)/(x^(1/10)+1)│+1/x^(1/10)-(1/2)/x^(1/5)+(1/3)/x^(3/10)-(1/4)/x^(2/5)+C
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