can someone help me solve
integral from 0 to 4 of
(x)/ (square root of x^2+9) dx
Just note that $\displaystyle x$ is (almost) the derivatived of $\displaystyle x^2+9$. Therefore you'd better substitute $\displaystyle z := x^2+9$ right away, which gives you that $\displaystyle dz = 2xdx$.
Plugging this into your integral allows you to convert it to the form $\displaystyle \tfrac{1}{2}\int \frac{1}{\sqrt{z}}\, dz$.
No, that's not quite correct, you should have $\displaystyle \tfrac{1}{2}\int\frac{1}{\sqrt{u}}\, du$
Finding the anti-derivative is the main problem, you have two choices as regards the limits of the integral: you can transform the limits along with the integrand:what do i do with the 0 to 4 part?
$\displaystyle \int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(9)}\frac{du}{\sqrt{u}} \, du = \Big[\sqrt{u}\Big]_{u=9}^{25} = \sqrt{25}-\sqrt{9}=2$
or you can do a back-substitution before you apply the old limits:
$\displaystyle \int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(4)}\frac{du}{\sqrt{u}} \, du = \Big[\sqrt{u}\Big]_{u=u(0)}^{u(4)} = \Big[\sqrt{x^2+9}\Big]_{x=0}^4=\ldots = 2$