Answer
$\displaystyle \sqrt{13}-\frac{3}{4}\ln(4+\sqrt{13})-\frac{1}{2}+\frac{3}{4}\ln 3$
Work Step by Step
Formula 39.
$\displaystyle \int\sqrt{u^{2}-a^{2}}du=\frac{u}{2}\sqrt{u^{2}-a^{2}}-\frac{a^{2}}{2}\ln|u+\sqrt{u^{2}-a^{2}}|+C$
---
$\displaystyle \int_{1}^{2}\sqrt{4x^{2}-3}dx= \qquad $
sub: $\left[\begin{array}{ll}
u =2x, & du =2dx\\
x=1\rightarrow u=2, & x=2\rightarrow u=4
\end{array}\right]$
$=\displaystyle \frac{1}{2}\int_{2}^{4}\sqrt{u^{2}-(\sqrt{3})^{2}}du = \qquad$ ... apply formula 39, $a=\sqrt{3}$
$=\displaystyle \frac{1}{2}\left[\frac{u}{2}\sqrt{u^{2}-(\sqrt{3})^{2}}-\frac{(\sqrt{3})^{2}}{2}\ln|u+\sqrt{u^{2}-(\sqrt{3})^{2}}|\right]_{2}^{4}$
$=\displaystyle \frac{1}{2}[\frac{4}{2}\sqrt{16-3}-\frac{3}{2}\ln|4+\sqrt{16-3}|]-\frac{1}{2}[\frac{2}{2}\sqrt{4-3}-\frac{3}{2}\ln|2+\sqrt{4-3}|]$
$=\displaystyle \frac{1}{2}[2\sqrt{13}-\frac{3}{2}\ln(4+\sqrt{13})]-\frac{1}{2}(1-\frac{3}{2}\ln 3)$
$=\displaystyle \sqrt{13}-\frac{3}{4}\ln(4+\sqrt{13})-\frac{1}{2}+\frac{3}{4}\ln 3$
