计算不定积分，谢谢大佬 求1/√(x^3)的不定积分，想要一个完整的分析过程，谢谢大...
1/\u6839\u53f7\u4e0b2-x^2\u7684\u4e0d\u5b9a\u79ef\u5206\u600e\u4e48\u6c42\uff0c\u8c22\u8c22\u5927\u4f6c\u4e71\u4e03\u516b\u7cdf\u7b54\u6848\u771f\u591a\u2026\u2026\u8be6\u7ec6\u8fc7\u7a0b\u5982\u56fert\u2026\u2026\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60\u89e3\u51b3\u95ee\u9898
\u7b54\u6848\u5982\u56fe\u6240\u793a\uff0c\u6ee1\u610f\u8bf7\u91c7\u7eb3\uff01
∴原式=∫[3/(x+2)-(2x+5)/(x²+3x+4)]dx=3ln丨x+2丨-ln(x²+3x+4)-(4/√7)arctan[(2x+3)/√7]+C。
第2题，∵x(x-1)=(x-1/2)²-1/4，∴设x-1/2=(1/2)secθ。原式=∫secθdθ=ln丨secθ+tanθ丨+C。∴原式=ln丨2x-1+2√(x²-x)丨+C。
第3题，原式=xln[x+√(x²+1)]-∫xdx/√(x²+1)=xln[x+√(x²+1)]-√(x²+1)+C。
供参考。
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