函数 $f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{1+x^2}}, x \leq 0 \\ (x+1) \cos x, x>0\end{array}\right.$ 的原函数为
A. $F(x)=\left\{\begin{array}{l}\ln \left(\sqrt{1+x^2}-x\right), x \leq 0 \\ (x+1) \cos x-\sin x, x>0\end{array}\right.$
B. $F(x)=\left\{\begin{array}{l}\ln \left(\sqrt{1+x^2}-x\right)+1, x \leq 0 \\ (x+1) \cos x-\sin x, x>0\end{array}\right.$
C. $F(x)=\left\{\begin{array}{l}\ln \left(\sqrt{1+x^2}-x\right), x \leq 0 \\ (x+1) \sin x+\cos x, x>0\end{array}\right.$
D. $F(x)=\left\{\begin{array}{l}\ln \left(\sqrt{1+x^2}+x\right)+1, x \leq 0 \\ (x+1) \sin x+\cos x, x>0\end{array}\right.$