## Calculus (3rd Edition)

$$\frac{3}{2}\left[\frac{(x-2) \sqrt{x^{2}-4 x+7}}{3}+\ln \left|\frac{(x-2)+\sqrt{x^{2}-4 x+7}}{\sqrt{3}}\right|\right]+C$$
Given $$\int \sqrt{x^{2}-4 x+7} d x$$ Since \begin{align*} x^{2}-4 x+7&=(x-2)^{2}-4 +7\\ &=(x-2)^{2}+3 \end{align*} Let $$x-2=\sqrt{3}\tan u\ \ \ \ \ \ dx=\sqrt{3}\sec^2 udu$$ then \begin{align*} \int \sqrt{x^{2}-4 x+7} d x&=\int \sqrt{(x-2)^{2}+3} d x\\ &=\sqrt{3}\int \sqrt{3\tan^{2}u+3} \sec^2 udu \\\\ &=3\int \sec^3udu \end{align*} Use $$\int \sec ^{n} u d u=\frac{1}{n-1} \tan u \sec ^{n-2} u+\frac{n-2}{n-1} \int \sec ^{n-2} u d u$$ then $$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C$$ Hence \begin{align*} \int \sqrt{x^{2}-4 x+7} d x &=3\int \sec^3udu\\ &= \frac{3}{2} \sec u \tan u+\frac{3}{2} \ln |\sec u+\tan u|+C\\ &=\frac{3}{2}\left[\frac{u \sqrt{u^{2}+3}}{3}+\ln \left|\frac{u+\sqrt{u^{2}+3}}{\sqrt{3}}\right|\right]+C\\ &=\frac{3}{2}\left[\frac{(x-2) \sqrt{x^{2}-4 x+7}}{3}+\ln \left|\frac{(x-2)+\sqrt{x^{2}-4 x+7}}{\sqrt{3}}\right|\right]+C \end{align*}