$\displaystyle \int \limits_0^{15} \dfrac{dx}{\sqrt{1+x}} =$

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Using substitution:

$$u=1+x$$ $$\frac{du}{dx}=1$$ $$\int \limits_0^{15} \dfrac{dx}{\sqrt{1+x}} = \int \limits_1^{16} \dfrac{du}{\sqrt{u}}$$

We can rewrite the expression to make it easier to integrate:

$$\int \limits_1^{16} \dfrac{du}{\sqrt{u}} = \int \limits_1^{16} u^{-\frac{1}{2}} du$$ $$= \left[ 2u^{\frac{1}{2}} \right]_1^{16} = \Big[ 2\sqrt{u} \Big]_1^{16}$$ $$=2(\sqrt{16}-\sqrt{1})$$ $$=2(4-1)$$ $$=\boxed{6}$$