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

Summary Submit Skip Question

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:

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