# How do you find int (x^2 + 1)/ (2x^3 - 5x^2 + x + 2)dx using partial fractions?

Dec 25, 2016

The answer is =-2/3ln(∣x-1∣)+ln(∣x-2∣)+1/6ln(∣2x+1∣)+C

#### Explanation:

Let $f \left(x\right) = 2 {x}^{3} - 5 {x}^{2} + x + 2$

$f \left(1\right) = 2 - 5 + 1 + 2 = 0$

Therefore, $\left(x - 1\right)$ is a factor

$f \left(2\right) = 16 - 20 + 2 + 2 = 0$

Therefore, $\left(x - 2\right)$ is a factor

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

To find the last factor, we do a long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - 5 {x}^{2} + x + 2$$\textcolor{w h i t e}{a a a a}$∣${x}^{2} - 3 x + 2$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - 6 {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a a a a}$∣$2 x + 1$

$\textcolor{w h i t e}{a a a a a a}$$0 + {x}^{2} - 3 x + 2$

$\textcolor{w h i t e}{a a a a a a a a}$$+ {x}^{2} - 3 x + 2$

$\textcolor{w h i t e}{a a a a a a a a a a}$$+ 0 - 0 + 0$

We can perform the decomposition into partial fractions

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

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

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

Therefore,

x^2+1=A(x-2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x-2))

Let $x = 1$, $\implies$, $2 = - 3 A$, $\implies$, $A = - \frac{2}{3}$

Let $x = 2$, $\implies$, $5 = 5 B$, $\implies$, $B = 1$

Let $x = - \frac{1}{2}$, $\implies$, $\frac{5}{4} = \frac{15}{4} C$, $\implies$, $C = \frac{1}{3}$

So,

$\frac{{x}^{2} + 1}{2 {x}^{3} - 5 {x}^{2} + x + 2} = \frac{- \frac{2}{3}}{x - 1} + \frac{1}{x - 2} + \frac{\frac{1}{3}}{2 x + 1}$

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

=-2/3ln(∣x-1∣)+ln(∣x-2∣)+1/6ln(∣2x+1∣)+C