$\displaystyle
\int_{1}^4 3/(2x + 1)^2\, dx
$
Pull out the 3 (because it's a constant and you can do that ), and then try u-substitution with u being (2x + 1)
Notice that this is
$\displaystyle \int_1^4{3(2x + 1)^{-2}\,dx}$
$\displaystyle = 3\int_1^4{(2x + 1)^{-2}\,dx}$
$\displaystyle = \frac{3}{2}\int_1^4{2(2x + 1)^{-2}\,dx}$.
You would solve this using a u-substitution.
Let $\displaystyle u = 2x + 1$ so that $\displaystyle \frac{du}{dx} = 2$
So the integral becomes
$\displaystyle \frac{3}{2}\int_{x = 1}^{x = 4}{u^{-2}\,\frac{du}{dx}\,dx}$
$\displaystyle = \frac{3}{2}\int_{x = 1}^{x = 4}{u^{-2}\,du}$
$\displaystyle = \frac{3}{2}\left[-u^{-1}\right]_{x = 1}^{x = 4}$
$\displaystyle = \frac{3}{2}\left[-\frac{1}{2x + 1}\right]_1^4$
$\displaystyle = \frac{3}{2}\left[\left(-\frac{1}{9}\right) - \left(-\frac{1}{3}\right)\right]$
$\displaystyle = \frac{3}{2}\left(\frac{1}{3} - \frac{1}{9}\right)$
$\displaystyle = \frac{3}{2}\left(\frac{2}{9}\right)$
$\displaystyle = \frac{1}{3}$.