Question: $$\int \sqrt{x}\log\_2(x) \, dx$$

Integral of Square root of x times log base 2 of x dx.

Answer: 
First lets change the base of the logarithm
$$\log\_2(x)=\frac{\log\_e(x)}{\log\_e(2)}=\frac{\ln(x)}{\ln(2)}$$ 
So your integral can be reformulated as 
$$\int\sqrt{x}\log\_2(x)\,dx=\int\sqrt{x}\frac{\ln(x)}{\ln(2)}\,dx=\frac{1}{\ln(2)}\int\sqrt{x}\ln(x)\,dx$$
Integrate by parts
$$\frac{1}{\ln(2)}\int\sqrt{x}\ln(x)\,dx=\frac{1}{\ln(2)}[\frac{2}{3}x^{3/2}\ln(x)-\frac{2}{3}\int x^{3/2}\cdot\frac{1}{x}\,dx]=\frac{1}{\ln(2)}[\frac{2}{3}x^{3/2}\ln(x)-\frac{4}{9}x^{3/2}+c]$$
If $u=\log\_2 x$ then $du= \dfrac {dx} {x\log\_e 2}$. If $dv=\sqrt{x}\,dx$ then $v = \dfrac 2 3 x^{3/2}$. So you get
\begin{align}
& \int u\,dv=vu-\int v\, du = \frac 2 3 x^{3/2}\log\_2 x - \int\frac 2 3 x^{3/2}\frac{dx}{x\log\_e 2} \\[8pt]
= {} & \frac 2 3 x^{3/2}\log\_2 x - \frac{2}{3\log\_e 2} \int x^{3/2}\frac {dx} x \\[8pt]
= {} & \frac 2 3 x^{3/2}\log\_2 x - \frac{2}{3\log\_e 2} \int x^{1/2}\, dx= \cdots
\end{align}
Here I used the fact that $x^{3/2}\dfrac 1 x= x^{1/2}$. Strangely, I have found that simple point to be a place where students get stuck after getting everything else right.