1. Let $I=\int_{-1}^{2}\ln\left(x+\sqrt{x^2+1}\right)dx.$

Recall the standard identity: $\operatorname{arsinh}(x)=\ln\left(x+\sqrt{x^2+1}\right).$ So, $I=\int_{-1}^{2}\operatorname{arsinh}(x)\,dx.$

2. Integrate by parts (or use the standard formula): $\int \operatorname{arsinh}(x)\,dx = x\operatorname{arsinh}(x)-\sqrt{x^2+1}+C.$

Since $\operatorname{arsinh}(x)=\ln\left(x+\sqrt{x^2+1}\right),$ we get $I=\left[x\ln\left(x+\sqrt{x^2+1}\right)-\sqrt{x^2+1}\right]_{-1}^{2}.$

3. Evaluate at the limits.

At $x=2$: $\sqrt{x^2+1}=\sqrt{5},$ $x+\sqrt{x^2+1}=2+\sqrt{5}.$ Hence, $F(2)=2\ln(2+\sqrt{5})-\sqrt{5}.$

At $x=-1$: $\sqrt{x^2+1}=\sqrt{2},$ $x+\sqrt{x^2+1}=-1+\sqrt{2}=\sqrt{2}-1.$ Hence, $F(-1)=-\ln(\sqrt{2}-1)-\sqrt{2}.$

Therefore, $I=F(2)-F(-1)$ $=2\ln(2+\sqrt{5})-\sqrt{5}+\ln(\sqrt{2}-1)+\sqrt{2}.$ So, $I=\sqrt{2}-\sqrt{5}+2\ln(2+\sqrt{5})+\ln(\sqrt{2}-1).$

4. Simplify the logarithmic part.

First, $2\ln(2+\sqrt{5})=\ln(2+\sqrt{5})^2=\ln(9+4\sqrt{5}),$ because $(2+\sqrt{5})^2=4+5+4\sqrt{5}=9+4\sqrt{5}.$

Also, $\sqrt{2}-1=\frac{1}{1+\sqrt{2}},$ so $\ln(\sqrt{2}-1)= -\ln(1+\sqrt{2})=\ln\left(\frac{1}{1+\sqrt{2}}\right).$

Thus,