I rewrote this as
= (2x + 1)(x + 1)^4 is that right??? then i just combine them and integrate after??????????????????
You need to rewrite them as $\displaystyle \int (2x+1)\cdot (x+4)^{-\frac{1}{2}}\text{ dx}$ and use integration by parts, so choose $\displaystyle u(x)=(x+4)^{-\frac{1}{2}}$ and $\displaystyle v(x)=2x+1$, then
$\displaystyle \int u(x)\cdot v(x) \text{ dx}=v(x)\cdot U(x)-\int v'(x)\cdot U(x)\text{ dx}$,
where $\displaystyle v'(x)$ is the differential of $\displaystyle v(x)$ and $\displaystyle U(x)$ is the antiderivative of $\displaystyle u(x)$.
Feel free to ask, if anything I just wrote need explanation
We know that $\displaystyle \int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$
More compactly:
$\displaystyle \int u \, dv=uv-\int v \, du.\!$
For $\displaystyle \int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx $ Let $\displaystyle dv=\frac{1}{\sqrt{x+4}}$ and $\displaystyle u=2x+1$
$\displaystyle \therefore v = 2\sqrt{x+4}\;\; \text{ and } du = 2\;\;dx$
where $\displaystyle v$ is the antiderivative of $\displaystyle dv$ and $\displaystyle du$ is the derivative of $\displaystyle u$
$\displaystyle \begin{align*}\int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx =& (2x+1)2\sqrt{x+4} - \int 4\sqrt{x+4} \;\;dx\\ =& 2(2x+1)\sqrt{x+4} -\frac{8}{3}(x+4)^{\frac{3}{2}}+C\\ =& \frac{2}{3}\sqrt{x+4}(2x-13)+C.......\text{[After simplification]}\end{align*}$
$\displaystyle \therefore\int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx = \frac{2}{3}\sqrt{x+4}(2x-13)+ C$
Check:
http://www.wolframalpha.com/input/?i=integration+(2*x%2B1)(x%2B4)^(-1%2F2)+dx