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

Oct 29, 2016

#### Explanation:

Expand:

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

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

Let x = -1 to make B and C disappear:

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

$- 1 = 4 A$

$A = - \frac{1}{4}$

${x}^{2} - 2 = - \frac{1}{4} \left({x}^{2} + 3\right) + \left(B x + C\right) \left(x + 1\right)$

Let x = 0 to make B disappear:

$- 2 = - \frac{1}{4} \left(3\right) + \left(C\right) \left(1\right)$

$C = - \frac{5}{4}$

${x}^{2} - 2 = - \frac{1}{4} \left({x}^{2} + 3\right) + \left(B x - \frac{5}{4}\right) \left(x + 1\right)$

Let x = 1:

${1}^{2} - 2 = - \frac{1}{4} \left({1}^{2} + 3\right) + \left(B - \frac{5}{4}\right) \left(1 + 1\right)$

$- 1 = - 1 + \left(2 B - \frac{5}{2}\right)$

$B = \frac{5}{4}$

$\int \frac{{x}^{2} - 2}{\left(x + 1\right) \left({x}^{2} + 3\right)} \mathrm{dx} = - \frac{1}{4} \int \frac{1}{x + 1} \mathrm{dx} + \frac{5}{4} \int \frac{x}{{x}^{2} + 3} \mathrm{dx} - \frac{5}{4} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

$\int \frac{{x}^{2} - 2}{\left(x + 1\right) \left({x}^{2} + 3\right)} \mathrm{dx} = - \frac{1}{4} \ln | x + 1 | + \frac{5}{4} \ln | {x}^{2} + 3 | - \frac{5 \sqrt{3}}{12} {\tan}^{-} 1 \left(\frac{x}{\sqrt{3}}\right)$