# How do you find int (x-1)/((x+2)(x^2+3))dx using partial fractions?

Apr 2, 2018

The answer is $= - \frac{3}{7} \ln \left(| x + 2 |\right) + \frac{3}{14} \ln \left({x}^{2} + 3\right) + \frac{1}{7 \sqrt{3}} \arctan \left(\frac{x}{\sqrt{3}}\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{x - 1}{\left(x + 2\right) \left({x}^{2} + 3\right)} = \frac{A}{x + 2} + \frac{B x + C}{{x}^{2} + 3}$

$= \frac{A \left({x}^{2} + 3\right) + \left(B x + C\right) \left(x + 2\right)}{\left(x + 2\right) \left({x}^{2} + 3\right)}$

The denominators are the same, compare the numerators

$x - 1 = A \left({x}^{2} + 3\right) + \left(B x + C\right) \left(x + 2\right)$

Let $x = - 2$, $\implies$, $- 3 = 7 A$, $\implies$, $A = - \frac{3}{7}$

Let $x = 0$, $\implies$, $- 1 = 3 A + 2 C$

$2 C = - 1 - 3 A = - 1 + \frac{9}{7} = \frac{2}{7}$

$C = \frac{1}{7}$

Coefficients of ${x}^{2}$

$0 = A + B$

$B = - A = \frac{3}{7}$

Therefore,

$\frac{x - 1}{\left(x + 2\right) \left({x}^{2} + 3\right)} = \frac{- \frac{3}{7}}{x + 2} + \frac{\left(\frac{3}{7}\right) x + \left(\frac{1}{7}\right)}{{x}^{2} + 3}$

So,

$\int \frac{\left(x - 1\right) \mathrm{dx}}{\left(x + 2\right) \left({x}^{2} + 3\right)} = \int \frac{- \frac{3}{7} \mathrm{dx}}{x + 2} + \int \frac{\left(\left(\frac{3}{7}\right) x + \left(\frac{1}{7}\right)\right) \mathrm{dx}}{{x}^{2} + 3}$

$= - \frac{3}{7} \ln \left(| x + 2 |\right) + \frac{3}{14} \int \frac{2 x \mathrm{dx}}{{x}^{2} + 3} + \frac{1}{7} \int \frac{1 \mathrm{dx}}{{x}^{2} + 3}$

$= - \frac{3}{7} \ln \left(| x + 2 |\right) + \frac{3}{14} \ln \left({x}^{2} + 3\right) + \frac{1}{21} \int \frac{\mathrm{dx}}{{\left(\frac{x}{\sqrt{3}}\right)}^{2} + 1}$

$= - \frac{3}{7} \ln \left(| x + 2 |\right) + \frac{3}{14} \ln \left({x}^{2} + 3\right) + \frac{1}{21} \cdot \sqrt{3} \arctan \left(\frac{x}{\sqrt{3}}\right) + C$

$= - \frac{3}{7} \ln \left(| x + 2 |\right) + \frac{3}{14} \ln \left({x}^{2} + 3\right) + \frac{1}{7 \sqrt{3}} \arctan \left(\frac{x}{\sqrt{3}}\right) + C$