If ∫1x1−x1+x dx=g(x)+c\displaystyle\int\frac1x\sqrt{\frac{1-x}{1+x}}\,dx=g(x)+c∫x11+x1−xdx=g(x)+c, g(1)=0g(1)=0g(1)=0, then g(12)g\left(\dfrac12\right)g(21) is equal to :
Official answer: A
Rewrite. 1x1−x1+x=1−xx1−x2=1x1−x2−11−x2\dfrac1x\sqrt{\dfrac{1-x}{1+x}}=\dfrac{1-x}{x\sqrt{1-x^2}}=\dfrac1{x\sqrt{1-x^2}}-\dfrac1{\sqrt{1-x^2}}x11+x1−x=x1−x21−x=x1−x21−1−x21.
Integrate (for 0<x<10<x<10<x<1): g(x)=−loge1+1−x2x−sin−1x+Cg(x)=-\log_e\dfrac{1+\sqrt{1-x^2}}x-\sin^{-1}x+Cg(x)=−logex1+1−x2−sin−1x+C; g(1)=0g(1)=0g(1)=0 gives C=π2C=\frac\pi2C=2π.
At 12\frac1221. g=−loge(2+3)+π3=loge3−13+1+π3g=-\log_e(2+\sqrt3)+\frac\pi3=\log_e\dfrac{\sqrt3-1}{\sqrt3+1}+\dfrac\pi3g=−loge(2+3)+3π=loge3+13−1+3π, since 3−13+1=2−3\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt33+13−1=2−3.
(A)