# How do you integrate f(x)=(x^2+x)/((3x^2-1)(x+7)) using partial fractions?

$\int \frac{{x}^{2} + x}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} \mathrm{dx}$
$= \frac{5}{219} \cdot \ln \left(3 {x}^{2} - 1\right) + \frac{\sqrt{3}}{146} \ln \left(\frac{x \sqrt{3} - 1}{x \sqrt{3} + 1}\right) + \frac{21}{73} \cdot \ln \left(x + 7\right) + {C}_{0}$

#### Explanation:

The solution is

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

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

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

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

Expanding the right side of the equation

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

Match the numerical coefficients of the left and right sides of the equation

$\frac{1 \cdot {x}^{2} + 1 \cdot x + 0 \cdot {x}^{0}}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} = \frac{\left(A + 3 C\right) {x}^{2} + \left(B + 7 A\right) x + \left(7 B - C\right) {x}^{0}}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)}$

The equations can now be established

$A + 3 C = 1 \text{ }$first equation
$B + 7 A = 1 \text{ }$second equation
$7 B - C = 0 \text{ }$third equation

from the third
$C = 7 B$
substitute this in the first
$A + 3 \left(7 B\right) = 1$
$A + 21 B = 1$
$A = 1 - 21 B \text{ }$fourth equation
from the second $B = 1 - 7 A$
and substitute in the fourth equation
$A = 1 - 21 B \text{ }$fourth equation
$A = 1 - 21 \left(1 - 7 A\right)$
$A = 1 - 21 + 147 A$
$A = \frac{20}{146} = \frac{10}{73}$

then $B = 1 - 7 A = 1 - 7 \left(\frac{10}{73}\right) = \frac{3}{73}$
$B = \frac{3}{73}$
$C = 7 B = 7 \left(\frac{3}{73}\right) = \frac{21}{73}$

Do the integration now

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

$\int \frac{{x}^{2} + x}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} \mathrm{dx} = \int \left(\frac{\left(\frac{10}{73}\right) x + \left(\frac{3}{73}\right)}{3 {x}^{2} - 1} + \frac{\frac{21}{73}}{x + 7}\right) \mathrm{dx}$

$\int \frac{{x}^{2} + x}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} \mathrm{dx}$
$= \int \frac{\frac{10}{73} x}{3 {x}^{2} - 1} \mathrm{dx} + \int \frac{\frac{3}{73}}{3 {x}^{2} - 1} \mathrm{dx} + \int \frac{\frac{21}{73}}{x + 7} \mathrm{dx}$

$\int \frac{{x}^{2} + x}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} \mathrm{dx}$
$= \frac{10}{6 \left(73\right)} \int \frac{6 x}{3 {x}^{2} - 1} \mathrm{dx} + \frac{3}{73} \int \frac{1}{3 {x}^{2} - 1} \mathrm{dx} + \frac{21}{73} \int \frac{1}{x + 7} \mathrm{dx}$

$\int \frac{{x}^{2} + x}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} \mathrm{dx}$
$= \frac{5}{219} \int \frac{6 x}{3 {x}^{2} - 1} \mathrm{dx} + \frac{\sqrt{3}}{73} \int \frac{\sqrt{3}}{{\left(\sqrt{3} x\right)}^{2} - {1}^{2}} \mathrm{dx} + \frac{21}{73} \int \frac{1}{x + 7} \mathrm{dx}$

Use the integration formulas

$\int \frac{\mathrm{du}}{u} = \ln u$ and $\int \frac{\mathrm{du}}{{u}^{2} - {a}^{2}} = \frac{1}{2 a} \ln \left(\frac{u - a}{u + a}\right)$

$\int \frac{{x}^{2} + x}{\left(3 {x}^{2} - 1\right) \left(x + 7\right)} \mathrm{dx}$
$= \frac{5}{219} \cdot \ln \left(3 {x}^{2} - 1\right) + \frac{\sqrt{3}}{146} \ln \left(\frac{x \sqrt{3} - 1}{x \sqrt{3} + 1}\right) + \frac{21}{73} \cdot \ln \left(x + 7\right) + {C}_{0}$

God bless....I hope the explanation is useful.