# How do you integrate int (4x)/sqrt(x^2-14x+53)dx using trigonometric substitution?

Jan 20, 2018

The answer is $= 4 \sqrt{{x}^{2} - 4 x + 53} + 28 \ln \left(\frac{x - 7}{2} + \sqrt{1 + {\left(x - 7\right)}^{2} / 4}\right) + C$

#### Explanation:

Let's rewrite the function

$\frac{4 x}{\sqrt{{x}^{2} - 14 x + 53}}$

$= 2 \left(\frac{2 x - 14}{\sqrt{{x}^{2} - 14 x + 53}}\right) + 2 \left(\frac{14}{\sqrt{{x}^{2} - 14 x + 53}}\right)$

Therefore, there are $2$ integrals

$\int \frac{4 x \mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}} = 2 \int \left(\frac{\left(2 x - 14\right) \mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}}\right) + 2 \int \left(\frac{14 \mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}}\right)$

Perform the integrals separately,

$2 \int \left(\frac{\left(2 x - 14\right) \mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}}\right) = 4 \sqrt{{x}^{2} - 4 x + 53}$

$2 \int \left(\frac{14 \mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}}\right) = 28 \int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}}$

Complete the squares in the denominator

${x}^{2} - 14 x + 53 = {x}^{2} - 14 x + 49 + 53 - 49 = {\left(x - 7\right)}^{2} + 4 = 4 \left({\left(\frac{x - 7}{2}\right)}^{2} + 1\right)$

Let $u = \frac{x - 7}{2}$, $\implies$, $\mathrm{du} = \frac{\mathrm{dx}}{2}$

$28 \int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}} = 28 \int \frac{\mathrm{dx}}{\sqrt{4 \left({\left(\frac{x - 7}{2}\right)}^{2} + 1\right)}}$

$= 14 \int \frac{2 \mathrm{du}}{\sqrt{{u}^{2} + 1}}$

$= 28 \int \frac{\mathrm{du}}{\sqrt{{u}^{2} + 1}}$

Let $u = \tan \theta$, $\implies$, $\mathrm{du} = {\sec}^{2} \theta d \theta$

$1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

$28 \int \frac{\mathrm{du}}{\sqrt{{u}^{2} + 1}} = 28 \int \frac{{\sec}^{2} \theta d \theta}{\sec \theta}$

$= 28 \int \sec \theta d \theta$

$= 28 \ln \left(\tan \theta + \sec \theta\right)$

$= 28 \ln \left(u + \sqrt{1 + {u}^{2}}\right)$

$= 28 \ln \left(\frac{x - 7}{2} + \sqrt{1 + {\left(x - 7\right)}^{2} / 4}\right)$

Finally,

$\int \frac{4 x \mathrm{dx}}{\sqrt{{x}^{2} - 14 x + 53}} = 4 \sqrt{{x}^{2} - 4 x + 53} + 28 \ln \left(\frac{x - 7}{2} + \sqrt{1 + {\left(x - 7\right)}^{2} / 4}\right) + C$