# How do you integrate int (4x)/sqrt(x^2-14x+65)dx using trigonometric substitution?

Mar 8, 2016

$4 \left(\sqrt{{x}^{2} - 14 x + 65} + 7 {\sinh}^{-} 1 \left(\frac{x - 7}{4}\right)\right) + C$

#### Explanation:

The two main tricks are:

1. See that: ${x}^{2} - 14 x + 65 = {\left(x - 7\right)}^{2} + {4}^{2}$
2. Use and abuse: ${\sinh}^{2} u + 1 = {\cosh}^{2} u$

Using 1. we choose our substitution like this:

• $x - 7 = 4 \sinh u$
• $\mathrm{dx} = 4 \cosh u \cdot \mathrm{du}$

To get the following:

${\left(x - 7\right)}^{2} + {4}^{2} \to {4}^{2} {\sinh}^{2} u + {4}^{2} = 16 {\cosh}^{2} u$

Which will simplify $\sqrt{{x}^{2} - 14 x + 65} \to 4 \cosh u$

We can proceed:
$\int \frac{4 x}{\sqrt{{x}^{2} - 14 x + 65}} \mathrm{dx} = \int \frac{4 \cdot \left(4 \sinh u + 7\right)}{4 \cosh u} \left(4 \cosh u\right) \mathrm{du}$
$= 4 \int \left(4 \sinh u + 7\right) \mathrm{du}$
$= 4 \cdot \left(4 \cosh u + 7 u\right) + c$

We can then substitute it back in x:

$4 \left(\sqrt{{x}^{2} - 14 x + 65} + 7 {\sinh}^{-} 1 \left(\frac{x - 7}{4}\right)\right) + C$