dx/dt = 1- sin[ ln(1+x )]
Replace the right hand side by the linear Taylor's approximation about x0=0 and solve the resulting equation. Determine the particular solution for the initial condition x(0)=0.
Taylor series for $\displaystyle \ln(1+x)$ and $\displaystyle \sin x$
$\displaystyle \ln (1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} \cdots$
$\displaystyle \sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} \cdots$
so to first order the taylor series would be
$\displaystyle \sin (\ln(1+x)) = x$