Answer
\[
\int_{-1}^{4} \frac{x}{\sqrt{5+X}} d x=\frac{8}{3}
\]
Work Step by Step
Let $u=5+x \Rightarrow d u=d x$
When $-1=x \Rightarrow u=4$
When $4=x \Rightarrow u=9$
\[
\begin{aligned}
\int_{-1}^{4} \frac{x}{\sqrt{5+X}} d x &=\int_{4}^{9} \frac{-5+u}{\sqrt{u}} d u \\
&=\int_{4}^{9}(-5+u) u^{-\frac{1}{2}} d u \\
&=\int_{4}^{9}\left(-5 u^{-\frac{1}{2}}+u^{\frac{1}{2}}\right) d u \\
&=\left.\left(-10 u^{\frac{1}{2}}+\frac{2}{3} u^{\frac{3}{2}}\right)\right|_{4} ^{9} \\
&=(-8+27)\frac{2}{3}-(-2+3)10 \\
&=\frac{8}{3}
\end{aligned}
\]
