Answer
$$
\int_{1}^{2} \sqrt{4 x^{2}-3} d x=\sqrt{13}-\frac{3}{4} \ln (4+\sqrt{13})-\frac{1}{2}+\frac{3}{4} \ln 3
$$
Work Step by Step
$$
\int_{1}^{2} \sqrt{4 x^{2}-3} d x
$$
If we look at the Table of Integrals , we see that the closest entry is number $39 $ with $a=\sqrt{3}:$
$$
\begin{aligned}
\int_{1}^{2} \sqrt{4 x^{2}-3} d x &=\frac{1}{2} \int_{2}^{4} \sqrt{u^{2}-(\sqrt{3})^{2}} d u \quad[u=2 x, d u=2 d x] \\
& \stackrel{39}{=} \frac{1}{2}\left[\frac{u}{2} \sqrt{u^{2}-(\sqrt{3})^{2}}-\frac{(\sqrt{3})^{2}}{2} \ln \left|u+\sqrt{u^{2}-(\sqrt{3})^{2}}\right|\right]_{2}^{4} \\
&=\frac{1}{2}\left[2 \sqrt{13}-\frac{3}{2} \ln (4+\sqrt{13})\right]-\frac{1}{2}\left(1-\frac{3}{2} \ln 3\right)\\
&=\sqrt{13}-\frac{3}{4} \ln (4+\sqrt{13})-\frac{1}{2}+\frac{3}{4} \ln 3
\end{aligned}
$$
