Question

求∫dx/(x√（1-x^2）)=

答案是：ln|(1-√(1-x^2))/x)|+c 求过程啊

Answer 1

∫ 1/[x√(1-x²)] dx
分子分母同乘以x
=∫ x/[x²√(1-x²)] dx
=(1/2)∫ 1/[x²√(1-x²)] d(x²)
令√(1-x²)=u，则x²=1-u²，d(x²)=-2udu
=(1/2)∫ 1/[(1-u²)u](-2u)du
=∫ 1/(u²-1) du
=(1/2)ln|(1-u)/(1+u)| + C
=(1/2)ln|(1-√(1-x²))/(1+√(1-x²))| + C
=(1/2)ln|(1-√(1-x²))²/x²| + C
=ln|(1-√(1-x²))/x| + C

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Answer 2

求不定积分∫dx/[x√(1-x²)]
解：令x=sinu，则dx=cosudu，代入原式得：
原式=∫cosudu/(sinucosu)=∫du/sinu=∫cscudu【分子分母同乘以(1-cosu)】
=∫[(1-cosu)cscu/(1-cosu)]du
=∫[(cscu-cosucscu)/(1-cosu)]du
=∫[(cscu-cotu)/(1-cosu)]du【分子分母再同乘以cscu】
=∫[(csc²u-cotucscu)/(cscu-cosucscu)]du
=∫[(-cotucscu+csc²u)/(cscu-cotu)]du
=∫d(cscu-cotu)/(cscu-cotu)
=ln∣cscu-cotu∣+C
=ln∣(1/x)-√(1-x²)/x∣+C
=ln∣[1-√(1-x²)]/x∣+C

Answer 3

√(1-x^2)=t.1-t^2=x^2, xdx=-tdt
∫dx/(x√（1-x^2))
=∫xdx/(x^2√（1-x^2))
=∫-tdt/t(1-t^2)
=∫dt/(t^2-1)
=(1/2)ln|(t-1)/(t+1)|+C
=(1/2)ln|(t-1)^2/(t^2-1)|+C
=ln|(√(1-x^2)-1)/x)|+C

Answer 4

啦啦啦 呦shi〜