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

Dec 3, 2015

$- \frac{5}{6} \ln | 2 x - 1 | + \frac{1}{3} \ln | x + 1 | + \ln | x - 1 | + c$

#### Explanation:

==============================

0) Preparations needed?

As the power of your denominator (${x}^{3}$) is bigger than the power of your numerator (${x}^{2}$) and your denominator is already factorized, you can start with the partial fraction decomposition right away!

==============================

1) Partial fraction decomposition

Your goal is to find $A$, $B$ and $C$ so that

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

... multiply both sides with $\left(2 x - 1\right) \left(x + 1\right) \left(x - 1\right)$....

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

... expand the terms at the right side...

$\iff {x}^{2} + 1 = A \cdot {x}^{2} - A + B \cdot 2 {x}^{2} - B \cdot 3 x + B + C \cdot 2 {x}^{2} + C \cdot x - C$

... "collect" the $\textcolor{red}{{x}^{2}}$ terms, the $\textcolor{b l u e}{x}$ terms and the $\textcolor{g r e e n}{\text{constant}}$ terms...

$\iff \textcolor{red}{{x}^{2}} + \textcolor{b l u e}{0 \cdot x} + \textcolor{g r e e n}{1} = \textcolor{red}{A {x}^{2}} + \textcolor{b l u e}{0 \cdot A x} \textcolor{g r e e n}{- A} + \textcolor{red}{2 B {x}^{2}} \textcolor{b l u e}{- 3 B x} + \textcolor{g r e e n}{B} + \textcolor{red}{2 C {x}^{2}} + \textcolor{b l u e}{C x} \textcolor{g r e e n}{- C}$

Now, to solve this equation, you can split it into three: one equation for ${x}^{2}$, one for $x$ and one for the constant terms:

{((I) color(white)(xxx) 1 = A + 2B + 2C color(white)(xxxxxxxxi) color(red)(x^2) " terms"), ((II) color(white)(xxi) 0 = -3B + C color(white)(xxxxxxxxxxx) color(blue)(x) " terms"), ((III) color(white)(xx) 1 = -A + B - C color(white)(xxxiii) color(green)("constant") " terms") :}

==============================

2) Solving linear equation system

To solve this system of linear equations, you can e.g. transform $\left(I I\right)$ as $C = 3 B$ and plug $3 B$ for $C$ into $\left(I\right)$ and $\left(I I I\right)$ obtaining $\left(I '\right)$ and $\left(I I I '\right)$:

{ ( (I') color(white)(xx) 1 = color(white)(xx) A + 8B ), ( (III') color(white)(x) 1 = - A - 2B ) :}

Adding those equation finally leads to $B = \frac{1}{3}$ and thus, the solution of the linear equation system is:

$A = - \frac{5}{3}$, $\textcolor{w h i t e}{\times} B = \frac{1}{3}$, $\textcolor{w h i t e}{\times} C = 1$.

==============================

3) Result of the partial fraction decomposition

This means that your original fraction can be transformed as follows:

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

==============================

4) Solving the integral

So, the last thing left to do is solving the integral!

$\int \frac{{x}^{2} + 1}{\left(2 x - 1\right) \left(x + 1\right) \left(x - 1\right)} \text{d} x$

$\textcolor{w h i t e}{\times \times \times \times} = \int \left(- \frac{5}{3} \cdot \frac{1}{2 x - 1} + \frac{1}{3} \cdot \frac{1}{x + 1} + \frac{1}{x - 1}\right) \text{d} x$

$\textcolor{w h i t e}{\times \times \times \times} = - \frac{5}{3} \int \frac{1}{2 x - 1} \text{d"x + 1/3 int 1/(x+1) "d"x + int 1/(x-1) "d} x$

$\textcolor{w h i t e}{\times \times \times \times} = - \frac{5}{3} \ln | 2 x - 1 | \cdot \frac{1}{2} + \frac{1}{3} \ln | x + 1 | + \ln | x - 1 | + c$