高数的不定积分的求解,我计算出来一半,实在是想不出来了
∫[xln(x+(1+x^2)^(1/2))]/[(1-x^2)^2]dx
=(1/2)∫[ln(x+(1+x^2)^(1/2)]/(1-x^2)^2d(x^2)
=(1/2)∫ln(x+(1+x^2)^(1/2)d(1/(1-x^2))
=ln[x+(1+x^2)^(1/2)]/[2(1-x^2)]-∫(2(1-x^2))^(-1)d[ln(x+(1+x^2))]
=ln[x+(1+x^2)^(1/2)]/[2(1-x^2)]+∫1/{2[(1+x^2)^(1/2)][x^2-1]}dx
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但是,对于求解∫1/{2[(1+x^2)^(1/2)][x^2-1]}dx,???????
[解决办法]
求到lz那一步之后令x=tant，化简一下就可以求出来了
