Answer
$$\frac{38}{3}$$
Work Step by Step
Given
$$ \int_{1}^{6} \sqrt{x+3} d x $$
Let
$$u= x+3 \ \ \ \ \Rightarrow \ \ \ du = dx $$
At $x=1 \to u= 4$ and at $x=6 \to u=9 $
Then
\begin{aligned} \int_{1}^{6} \sqrt{x+3} d x &=\int_{4}^{9} \sqrt{u} d u \\ &=\left.\frac{2}{3} u^{3 / 2}\right|_{4} ^{9} \\ &=\frac{2}{3}\left(9^{3 / 2}-4^{3 / 2}\right) \\ &=\frac{2}{3}(27-8)=\frac{2}{3}(19)\\
&=\frac{38}{3}
\end{aligned}
