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

Nov 18, 2017

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

=$\frac{1}{2} \cdot L n x$+$\frac{\sqrt{2}}{6} \cdot \arctan \left(\frac{x}{\sqrt{2}}\right)$+$\frac{1}{12} \cdot L n \left({x}^{2} + 2\right)$-$\frac{2}{3} \cdot L n \left(x + 1\right) + C$

#### Explanation:

I decomposed integrand into basic fractions,

$\frac{{x}^{2} + 1}{x \cdot \left(x + 1\right) \cdot \left({x}^{2} + 2\right)}$

=$\frac{A}{x} + \frac{B}{x + 1} + \frac{C x + D}{{x}^{2} + 2}$

After expanding denominator,

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

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

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

After equating coefficients,

$A + B + C = 0$, $A + C + D = 1$, $2 A + 2 B + D = 0$ and $2 A = 1$

From these equations, $A = \frac{1}{2} , B = - \frac{2}{3} , C = \frac{1}{6} \mathmr{and} D = \frac{1}{3}$

Thus,

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

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

=$\frac{1}{2} \cdot L n x - \frac{2}{3} \cdot L n \left(x + 1\right) + \frac{1}{12} \cdot \int \frac{\left(2 x + 4\right) \cdot \mathrm{dx}}{{x}^{2} + 2}$

=$\frac{1}{2} \cdot L n x - \frac{2}{3} \cdot L n \left(x + 1\right) + \frac{1}{12} \cdot \int \frac{2 x \cdot \mathrm{dx}}{{x}^{2} + 2} + \frac{1}{3} \cdot \int \frac{\mathrm{dx}}{{x}^{2} + 2}$

=$\frac{1}{2} \cdot L n x - \frac{2}{3} \cdot L n \left(x + 1\right) + \frac{1}{12} \cdot L n \left({x}^{2} + 2\right) + \frac{\sqrt{2}}{6} \cdot \int \frac{\sqrt{2} \cdot \mathrm{dx}}{{x}^{2} + 2}$

=$\frac{1}{2} \cdot L n x - \frac{2}{3} \cdot L n \left(x + 1\right) + \frac{1}{12} \cdot L n \left({x}^{2} + 2\right) + \frac{\sqrt{2}}{6} \cdot \arctan \left(\frac{x}{\sqrt{2}}\right) + C$