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

$\frac{1}{2}\left[sin^{-1}\frac{x}{a}+ln\sqrt{x+\sqrt{a^2-x^2}}+c\right]$ $\frac{1}{2}sin^{-1}\frac{x}{a}+ln(x+\sqrt{a^2-x^2})+c$ $\frac{1}{2}sin^{-1}\frac{x}{a}+ln(\sqrt{a+\sqrt{a^2-x^2}})+c$ none of these.

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

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.