## Calculus (3rd Edition)

$$t \sqrt{t^{2}+3}+3 \ln \left|\frac{\sqrt{t^{2}+3}+t}{\sqrt{3}}\right|+C$$
Given $$\sqrt{12+4 t^{2}} d t$$ Let $$t= \sqrt{3}\tan \theta \ \ \ dt= \sqrt{3}\sec^2 \theta d \theta$$ Then \begin{aligned} \int \sqrt{12+4 t^{2}} d t &=\int \sqrt{4\left(3+t^{2}\right)} d t \\ &=2 \int \sqrt{3+t^{2}} d t \\ &=2 \int \sqrt{3+(\sqrt{3} \tan \theta)^{2}} \cdot \sqrt{3} \sec ^{2} \theta d \theta \\ &=2 \int \sqrt{3\left(1+\tan ^{2} \theta\right)^{2}} \cdot \sqrt{3} \sec ^{2} \theta d \theta \\ &=2 \int 3 \sqrt{\sec ^{2} \theta} \cdot \sec ^{2} \theta d \theta \\ &=6 \int \sec \theta \cdot \sec ^{2} \theta d \theta \\ &=6 \int \sec ^{3} \theta d \theta \end{aligned} Use $$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$ Then \begin{aligned} \int \sqrt{12+4 t^{2}} d t &=6\left[\frac{\tan \theta \sec ^{3-2} \theta}{3-1}+\frac{3-2}{3-1} \int \sec ^{3-2} \theta d \theta\right]\\ &=6\left[\frac{\tan \theta \sec \theta}{2}+\frac{1}{2} \int \sec \theta d \theta\right]\\ &=6 \cdot \frac{1}{2}[\tan \theta \sec \theta+\ln |\sec \theta+\tan \theta|]+C\\ &=3 \tan \theta \sec \theta+3 \ln |\sec \theta+\tan \theta|+C\\ &=t \sqrt{t^{2}+3}+3 \ln \left|\frac{\sqrt{t^{2}+3}+t}{\sqrt{3}}\right|+C \end{aligned}