# How do you integrate int (2x-2)/(x^2 - 6x +10)^(1/2)dx using partial fractions?

Nov 22, 2015

$\left[2 \sqrt{{x}^{2} - 6 x + 10}\right] + \left[4 \arcsin h \left(x - 3\right)\right] + C$

#### Explanation:

$\int \frac{2 x - 2}{\sqrt{{x}^{2} - 6 x + 10}} \mathrm{dx}$

$\int \frac{2 x - 6 + 4}{\sqrt{{x}^{2} - 6 x + 10}} \mathrm{dx}$

$\int \frac{2 x - 6}{\sqrt{{x}^{2} - 6 x + 10}} \mathrm{dx} + 4 \int \frac{1}{\sqrt{{x}^{2} - 6 x + 10}} \mathrm{dx}$

For the first integral let's $u = {x}^{2} - 6 x + 10$

$\mathrm{du} = 2 x - 6 \mathrm{dx}$

we have $\int \frac{1}{\sqrt{u}} \mathrm{du}$

which is $\left[2 \sqrt{u}\right]$

complete the square for the second

${x}^{2} - 6 x + 10 = {x}^{2} - 6 x + 10 + 9 - 9 = {\left(x - 3\right)}^{2} + 1$

$\left[2 \sqrt{{x}^{2} - 6 x + 10}\right] + 4 \int \frac{1}{\sqrt{{\left(x - 3\right)}^{2} + 1}} \mathrm{dx}$

let's $t = x - 3$

$\mathrm{dt} = \mathrm{dx}$

$\left[2 \sqrt{{x}^{2} - 6 x + 10}\right] + 4 \int \frac{1}{\sqrt{{t}^{2} + 1}} \mathrm{dx}$

$\left[2 \sqrt{{x}^{2} - 6 x + 10}\right] + \left[4 \arcsin h \left(x - 3\right)\right] + C$