# How do you integrate int 1/sqrt(4x^2+16x+13)  using trigonometric substitution?

$\frac{1}{2} \setminus \ln | 2 \left(x + 2\right) + \setminus \sqrt{4 {x}^{2} + 16 x + 13} | + C$

#### Explanation:

$I = \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{4 {x}^{2} + 16 x + 13}}$

$= \setminus \int \setminus \frac{\mathrm{dx}}{2 \setminus \sqrt{{x}^{2} + 4 x + \setminus \frac{13}{4}}}$

$= \frac{1}{2} \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{{\left(x + 2\right)}^{2} - \setminus \frac{3}{4}}}$

Let $x + 2 = \setminus \frac{\setminus \sqrt{3}}{2} \setminus \sec \setminus \theta \setminus \implies \mathrm{dx} = \setminus \frac{\setminus \sqrt{3}}{2} \setminus \sec \setminus \theta \setminus \tan \setminus \theta d \setminus \theta$

$= \frac{1}{2} \setminus \int \setminus \frac{\setminus \frac{\setminus \sqrt{3}}{2} \setminus \sec \setminus \theta \setminus \tan \setminus \theta \setminus d \setminus \theta}{\setminus \sqrt{\frac{3}{4} \setminus {\sec}^{2} \setminus \theta - \setminus \frac{3}{4}}}$

$= \frac{1}{2} \setminus \int \setminus \frac{\setminus \sec \setminus \theta \setminus \tan \setminus \theta \setminus d \setminus \theta}{\setminus \tan \setminus \theta}$

$= \frac{1}{2} \setminus \int \setminus \sec \setminus \theta d \setminus \theta$

$= \frac{1}{2} \setminus \ln | \setminus \sec \setminus \theta + \setminus \tan \setminus \theta | + c$

$= \frac{1}{2} \setminus \ln | \setminus \frac{2}{\setminus \sqrt{3}} \left(x + 2\right) + \setminus \sqrt{\frac{4}{3} {\left(x + 2\right)}^{2} - 1} | + c$

$= \frac{1}{2} \setminus \ln | \left(x + 2\right) + \setminus \sqrt{{\left(x + 2\right)}^{2} - \frac{3}{4}} | + c '$

$= \frac{1}{2} \setminus \ln | 2 \left(x + 2\right) + \setminus \sqrt{4 {x}^{2} + 16 x + 13} | + C$