# How do you integrate int (3-2x)/(x(x²+3)) using partial fractions?

Dec 19, 2017

$L n x - \frac{1}{2} L n \left({x}^{2} + 3\right) - \frac{2 \sqrt{3}}{3} \cdot \arctan \left(\frac{x}{\sqrt{3}}\right) + C$

#### Explanation:

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

I decomposed integrand into basic fractions

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

=$\frac{A}{x} + \frac{B x + C}{{x}^{2} + 3}$

After expanding denominator,

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

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

After equating coefficients, $A + B = 0$, $C = - 2$ and $3 A = 3$

From them, $A = 1$, $B = - 1$ and $C = - 2$

Thus,

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

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

=$L n x - \int \frac{x}{{x}^{2} + 3} \cdot \mathrm{dx} - \int \frac{2}{{x}^{2} + 3} \cdot \mathrm{dx}$

=$L n x - \frac{1}{2} L n \left({x}^{2} + 3\right) - \frac{2 \sqrt{3}}{3} \cdot \arctan \left(\frac{x}{\sqrt{3}}\right) + C$