1. We are asked to find the integral of $\ln(1-\sqrt{x})\,dx$. 2. The integral is $\int \ln(1-\sqrt{x})\,dx$. 3. Use substitution: let $t=\sqrt{x}$, so $x=t^2$ and $dx=2t\,dt$. 4. Rewrite the integral in terms of $t$: $$\int \ln(1-t) \cdot 2t\,dt = 2 \int t \ln(1-t)\,dt$$ 5. Use integration by parts: let $u=\ln(1-t)$ and $dv=t\,dt$. Then $du=\frac{-1}{1-t} dt$ and $v=\frac{t^2}{2}$. 6. Integration by parts formula: $$\int u\,dv = uv - \int v\,du$$ Apply it: $$2 \left( \frac{t^2}{2} \ln(1-t) - \int \frac{t^2}{2} \cdot \frac{-1}{1-t} dt \right) = t^2 \ln(1-t) + \int \frac{t^2}{1-t} dt$$ 7. Simplify the integral: $$\int \frac{t^2}{1-t} dt = \int \frac{t^2 - 1 + 1}{1-t} dt = \int \frac{(t^2 -1) + 1}{1-t} dt = \int \frac{(t-1)(t+1) + 1}{1-t} dt$$ 8. Rewrite denominator: $$\frac{(t-1)(t+1) + 1}{1-t} = \frac{-(1-t)(t+1) + 1}{1-t} = - (t+1) + \frac{1}{1-t}$$ 9. So the integral becomes: $$\int \left(- (t+1) + \frac{1}{1-t} \right) dt = \int - (t+1) dt + \int \frac{1}{1-t} dt$$ 10. Integrate each term: $$- \int (t+1) dt = - \left( \frac{t^2}{2} + t \right)$$ $$\int \frac{1}{1-t} dt = - \ln|1-t|$$ 11. Combine results: $$- \frac{t^2}{2} - t - \ln|1-t| + C$$ 12. Substitute back into the expression from step 6: $$t^2 \ln(1-t) + \left(- \frac{t^2}{2} - t - \ln|1-t| \right) + C$$ 13. Finally, substitute $t=\sqrt{x}$: $$x \ln(1-\sqrt{x}) - \frac{x}{2} - \sqrt{x} - \ln|1-\sqrt{x}| + C$$ Answer: $$\int \ln(1-\sqrt{x})\,dx = x \ln(1-\sqrt{x}) - \frac{x}{2} - \sqrt{x} - \ln|1-\sqrt{x}| + C$$