We need to evaluate $$\int_{-3}^{1}\sqrt{(x^2+6x+13)}dx$$

Completing the square, $$x^2+6x+13=(x+3)^2+4$$

\since there is $$+$$ sign completing the square, we need to use the identity $$\sinh^2A+1=\cosh^2A$$

Now make the substituition $$x+3=2\sinh\theta$$

$$\Rightarrow \dfrac{dx}{d\theta}=2\cosh\:\theta$$

When $$x=-3,\sinh\:\theta=0\Rightarrow \theta=0$$

When $$x=1,\sinh\:\theta=2\Rightarrow \theta=\sinh^{-1}2=ln(2+\sqrt{5})$$

$$\therefore \int_{-3}^{1}\sqrt{(x^2+6x+13)}dx=\int_{0}^{\sinh^{-1}2}\sqrt{4\sinh^2\theta+4}\cdot 2\cosh\:\theta\:d\theta$$

$$=\int_{0}^{\sinh^{-1}2}4\cosh^2\theta\:d\theta$$

U\sing the identity $$\cosh\:2A=2\cosh^2A-1$$ we get

$$=\int_{0}^{\sinh^{-1}2}(2+2\cosh\:2\theta)d\theta$$

$$=\left[2\theta+\sinh\:2\theta\right]_{0}^{\sinh^{-1}2}$$

$$=2[\theta+\sinh\:\theta \cosh\:\theta]_{0}^{\sinh^{-1}2}$$

$$=2\left(\sinh^{-1}2+2\sqrt{1+2^2}\right)$$

$$=2ln(2+\sqrt{5})+4\sqrt{5}$$