i got (integral) 1/(u^2 +7/4) du
the answer is [2/sq.rt.7]( arctan (2u/sq.rt.7))
arctan u is the integral of (1/1+u^2)
Remember that in general $\displaystyle \int\frac{1}{x^2+a^2}dx=\int\frac{dx}{a^2\left(\fr ac{x^2}{a^2}+1\right)}=\frac{1}{a^2}\int\frac{dx}{ \left(\frac{x}{a}\right)^2+1}$
In this case, your $\displaystyle a^2=\frac{7}{4}$
So you're gonna need to use a substitution $\displaystyle u=\frac{2x}{\sqrt{7}}$.
EDIT: Sorry I used a different variable. I forgot that you're using u instead of x lol.