$∫\frac{dx}{x+\sqrt{a^2-x^2}}$ is equal to

$\frac{1}{2}\left[sin^{-1}\frac{x}{a}+ln\sqrt{x+\sqrt{a^2-x^2}}+c\right]$

Here $I=∫\frac{cosθ}{sinθ+cosθ}dθ$  (x = a sin θ, dx = a cos θ dθ)

$=\frac{1}{2}∫dθ+\frac{1}{2}∫\frac{cosθ-sinθ}{sinθ+cosθ}dθ=\frac{1}{2}θ+\frac{1}{2}ln(sinθ+cosθ)+c$.

$=\frac{1}{2}\left[sin^{-1}\frac{x}{a}+ln\sqrt{x+\sqrt{a^2-x^2}}+c\right]$

Hence (1) is the correct answer.