The value of  $\int \frac{\left({\tan }^{-1}(\sin x+1)\right)\cos x}{(3+2\sin x-{\cos }^{2}x)}dx$  is (where c is the constant integration)

$\frac{{\left({\tan }^{-1}(\sin x+1)\right)}^2}{2}+c$

Let,  $I=\int \frac{{\tan }^{-1}(\sin x+1)\cos x}{3+2\sin x-(1-{\sin }^{2}x)}dx$

Substituting, $1+\sin x=t\Rightarrow \cos dx=dt$

$I=\int \frac{{\tan }^{-1}\left(t\right)dt}{2+{(t-1)}^2+2(t-1)}=\int \frac{{\tan }^{-1}\left(t\right)dt}{1+t^2}=\frac{{\left({\tan }^{-1}\left(t\right)\right)}^2}{2}+c$

$I=\frac{{\left({\tan }^{-1}(\sin x+1)\right)}^2}{2}+c$