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

$\int \frac{3 - x}{\left({x}^{2} + 3\right) \left(x + 3\right)} \mathrm{dx}$
$= - \frac{1}{4} \ln \left({x}^{2} + 3\right) + \frac{\sqrt{3}}{6} {\tan}^{-} 1 \left(\frac{\sqrt{3} \cdot x}{3}\right) + \frac{1}{2} \ln \left(x + 3\right) + {C}_{0}$

#### Explanation:

We start with partial fraction procedure with the unknowns A, B, C

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

LCD$= \left({x}^{2} + 3\right) \left(x + 3\right)$

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

Transform the numerator so that the numerical coefficients match for both left and right side of the equation

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

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

We now have the equations to solve for the values of A,B, C

$A + C = 0$
$3 A + B = - 1$
$3 B + 3 C = 3$

Simultaneous solution of these equations result to
$A = - \frac{1}{2}$ and $B = \frac{1}{2}$ and $C = \frac{1}{2}$

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

We can now integrate

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

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

$\int \frac{3 - x}{\left({x}^{2} + 3\right) \left(x + 3\right)} \mathrm{dx} = - \frac{1}{4} \ln \left({x}^{2} + 3\right) + \frac{\sqrt{3}}{6} {\tan}^{-} 1 \left(\frac{\sqrt{3} \cdot x}{3}\right) + \frac{1}{2} \ln \left(x + 3\right) + {C}_{0}$

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