题目
曲线 y=ln(1-x^2) 上 0 leq x leq (1)/(2) 一段弧长 s=A. int_(0)^(1)/(2) sqrt(1+((1)/(1-x^2))^2) dxB. int_(0)^(1)/(2) (1+x^2)/(1-x^2) dxC. int_(0)^(1)/(2) sqrt(1+(-2x)/(1-x^2)) dxD. int_(0)^(1)/(2) sqrt(1+[ln(1-x^2)]^2) dx
曲线 $y=\ln(1-x^2)$ 上 $0 \leq x \leq \frac{1}{2}$ 一段弧长 $s=$
A. $\int_{0}^{\frac{1}{2}} \sqrt{1+(\frac{1}{1-x^2})^2} dx$
B. $\int_{0}^{\frac{1}{2}} \frac{1+x^2}{1-x^2} dx$
C. $\int_{0}^{\frac{1}{2}} \sqrt{1+\frac{-2x}{1-x^2}} dx$
D. $\int_{0}^{\frac{1}{2}} \sqrt{1+[\ln(1-x^2)]^2} dx$
题目解答
答案
B. $\int_{0}^{\frac{1}{2}} \frac{1+x^2}{1-x^2} dx$