How do you integrate int (2x-2)/ (x^2 - 4x + 13)dx using partial fractions?

Mar 7, 2018

$I = \ln | {x}^{2} - 4 x + 13 | + \frac{2}{3} \cdot {\tan}^{_ 1} \left(\frac{x - 2}{3}\right) + C$

Explanation:

$\textcolor{red}{\left(1\right) \int \frac{{F}^{'} \left(x\right)}{F \left(x\right)} \mathrm{dx} = \ln | F \left(x\right) | + c}$
$\textcolor{red}{\left(2\right) \int \frac{1}{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{1}{a} \cdot {\tan}^{1} \left(\frac{x}{a}\right) + c}$
Here,
$\int \frac{2 x - 2}{{x}^{2} - 4 x + 13} \mathrm{dx} = \int \frac{2 x - 4 + 2}{\left({x}^{2} - 4 x + 4 + 9\right)} \mathrm{dx} = \int \frac{2 \left(x - 2\right) + 2}{{\left(x - 2\right)}^{2} + 9} \mathrm{dx}$
take, $x - 2 = t \Rightarrow \mathrm{dx} = \mathrm{dt}$
$I = \int \frac{2 t + 2}{{t}^{2} + 9} \mathrm{dt} = \int \frac{2 t}{{t}^{2} + 9} \mathrm{dt} + 2 \int \frac{1}{{t}^{2} + {3}^{2}} \mathrm{dt}$
$I = \int \frac{\frac{d}{\mathrm{dt}} \left({t}^{2} + 9\right)}{{t}^{2} + 9} \mathrm{dt} + 2 \cdot \left(\frac{1}{3}\right) \cdot {\tan}^{- 1} \left(\frac{t}{3}\right) + c$
$I = \ln | {t}^{2} + 9 | + \frac{2}{3} \cdot {\tan}^{- 1} \left(\frac{t}{3}\right) + C$
$I = \ln | {\left(x - 2\right)}^{2} + 9 | + \frac{2}{3} \cdot {\tan}^{- 1} \left(\frac{x - 2}{3}\right) + C$
$I = \ln | {x}^{2} - 4 x + 13 | + \frac{2}{3} \cdot {\tan}^{_ 1} \left(\frac{x - 2}{3}\right) + C$