Answer
$$
f^{\prime}(x)=\frac{(x+1)}{\sqrt{x}}, \quad f(1)=5
$$
The required function is
$$
f(x)=\frac{2}{3} x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}+2
$$
Work Step by Step
$$
f^{\prime}(x)=\frac{(x+1)}{\sqrt{x}}, \quad f(1)=5
$$
can be written as the following:
$$
f^{\prime}(x)=\frac{(x+1)}{\sqrt{x}}=x^{-\frac{1}{2}}(x+1)=x^{\frac{1}{2}}+ x^{-\frac{1}{2}}
$$
The general anti-derivative of $
f^{\prime}(x)=\frac{(x+1)}{\sqrt{x}}
$ is
$$
f(x)=\frac{2}{3} x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}+C
$$
To determine C we use the fact that $f(1)=5$:
$$
f(1)=\frac{2}{3}(1)^{\frac{3}{2}}+ 2(1)^{\frac{1}{2}}+C=5
$$
$ \Rightarrow $
$$
3+C=5\quad \Rightarrow \quad C=2,
$$
so
$$
f(x)=\frac{2}{3} x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}+2
$$
