Answer
$\displaystyle \int_{0}^{3}\dfrac{1}{\sqrt{1+x}}dx = 2$
Work Step by Step
To evaluate the integral $\displaystyle \int_{0}^{3}\dfrac{1}{\sqrt{1+x}}dx$, let $u = 1+x$.
Then we obtain $u = 1 + x \longrightarrow du = dx $
Before substituting, determine the new upper and lower limits of integration
Lower limit: When $x = 0$, $u = 1+0 = 1$
Upper limit: When $x = 3$, $u = 1+3= 4$
$\displaystyle \int_{0}^{3}\dfrac{1}{\sqrt{1+x}}dx = \displaystyle \int_{1}^{4}\dfrac{1}{\sqrt{u}}du = \displaystyle \int_{1}^{4}\dfrac{1}{u^{1/2}}du = \displaystyle \int_{1}^{4}u^{-1/2}du = \dfrac{u^{1/2}}{\frac{1}{2}} \Bigg\vert_{1}^{4} = 2u^{1/2} \Bigg\vert_{1}^{4} = 2 \big[ 4^{1/2} - 1^{1/2}\big] = 2 \big[ 2 - 1\big] = 2 \big[ 1\big] = 2$
