# How do you integrate int (2x+1) /( (x-2)(x^2+4)) using partial fractions?

Sep 30, 2017

$\frac{\left(5 - \ln | x - 2 |\right) - \left(3 \cdot {\tan}^{-} 1 \left(\frac{x}{2}\right)\right)}{8} + \frac{5 \ln | {x}^{2} + 4 |}{16} + c$

#### Explanation:

1. First step in this problem is understanding that the use of partial fractions is valid since the numerator is a lesser degree of power than the denominator.

2. Apply Partial Fractions:
*Note that the denominator terms are Linear (left) and Quadratic (on the right), all this means is that they are terms that cannot be factorized further.
Linear form = $\frac{A}{x - 2}$
Quadratic form = $\frac{B x + C}{{x}^{2} + 4}$

Add these two terms together and set it equal to the Numerator.
$2 x + 1 = \frac{A}{x - 2} + \frac{B x + C}{{x}^{2} + 4}$

Simplify by multiplying A by the denomenator of the other fraction, and the same for B:
$2 x + 1 = A \left({x}^{2} + 4\right) + B x + C \left(x - 2\right)$

now distribute:
$2 x + 1 = A {x}^{2} + 4 A + B {x}^{2} - 2 B x + C x - 2 C$

*Note: The biggest thing students confuse about partial fractions is this next step, in the left hand side you should think of it as $0 {x}^{2} + 2 x + 1$ to understand that there are Zero ${x}^{2}$'s on the left.

$\textcolor{red}{0 {x}^{2}} + \textcolor{b l u e}{2 x} + \textcolor{g r e e n}{1} = \textcolor{red}{A {x}^{2}} + \textcolor{g r e e n}{4 A} + \textcolor{red}{B {x}^{2}} - \textcolor{b l u e}{2 B x} + \textcolor{b l u e}{C x} - \textcolor{g r e e n}{2 C}$

Collect like terms:
$\textcolor{red}{{x}^{2}} \Rightarrow$ $0 = A + B$
$\textcolor{b l u e}{x} \Rightarrow$ $2 = - 2 B + C$
$\textcolor{g r e e n}{\text{Integers}}$ $\Rightarrow$ $1 = 4 A - 2 C$

Solve by any method, here we will use substitution
Eq1: $0 = A + B \Rightarrow \left[A = - B\right] \Leftrightarrow \left[B = - A\right]$
Eq2: $2 = - 2 B + C \Rightarrow \left[2 + 2 B = C\right]$
Eq3: $1 = 4 A - 2 C \Rightarrow 1 = 2 \left(2 A - C\right) \Rightarrow \left[\frac{1}{2} = 2 A - C\right]$

Substitute in Eq1 and Eq2 into Eq3 to get:
$\frac{1}{2} = 2 \left(- B\right) - \left(2 + 2 B\right)$
Simplify:
$\frac{1}{2} = - 4 B - 2$

Solve for B:
$B = - \frac{5}{8}$
Use B to solve for A
$A = \frac{5}{8}$
Use B to solve for C
$2 + 2 \left(- \frac{5}{8}\right) = C$
$C = \frac{3}{4}$

Plug in A, B, and C values into original fraction:
$\frac{A}{x - 2} + \frac{B x + C}{{x}^{2} + 4}$

$\frac{\frac{5}{8}}{x - 2} + \frac{- \frac{5}{8} x + \frac{3}{4}}{{x}^{2} + 4}$

Simplify into:
$\frac{5}{8 \left(x - 2\right)} + \frac{- 5 x + 6}{8 \left({x}^{2} + 4\right)}$

New integral is now represented by:
$\int \frac{5}{8 \left(x - 2\right)} \mathrm{dx} + \int \frac{- 5 x + 6}{8 \left({x}^{2} + 4\right)} \mathrm{dx}$

Remove constants to prepare for integration:
$5 \cdot \int \frac{1}{8 \left(x - 2\right)} \mathrm{dx} + \frac{1}{8} \cdot \int \frac{- 5 x + 6}{\left({x}^{2} + 4\right)} \mathrm{dx}$

Break apart second integral (on the right) into two separate integrals (Mind the 1/8 multiplier) and remove the constants in the numerators

$\frac{5}{8} \cdot \int \frac{1}{\left(x - 2\right)} \mathrm{dx} + \frac{1}{8} \cdot \left(- 5 \cdot \int \frac{x}{{x}^{2} + 4} \mathrm{dx} + 6 \cdot \int \frac{1}{{x}^{2} + 4} \mathrm{dx}\right)$

First integral:
$\frac{5}{8} \int \frac{1}{x - 2} \mathrm{dx}$
$u = x - 2$
Result is: $\frac{5 \ln | x - 2 |}{8}$

Second Integral:
$5 \cdot \int \frac{x}{{x}^{2} + 4} \mathrm{dx}$
$u = {x}^{2} + 4$
Result Is: $\frac{- 5 \ln | {x}^{2} + 4 |}{2}$

Third Integral:
$6 \cdot \int \frac{1}{{x}^{2} + 4} \mathrm{dx}$
Apply inverse trig formula by identifying denominator as Arctan:
$\int \frac{1}{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{1}{a} \cdot {\tan}^{-} 1 \left(\frac{x}{a}\right)$
Where $\left[{a}^{2} = 4\right] \Rightarrow \left[a = 2\right]$
Result is: $3 {\tan}^{-} 1 \left(\frac{x}{2}\right)$

Plug these back into full equation and do not forget the $\frac{1}{8}$ multiplier for the second and third integrals.

$\frac{5 \cdot \ln | x - 2 |}{8} - \frac{1}{8} \left(\frac{- 5 \ln | {x}^{2} + 4 |}{2} + 3 {\tan}^{-} 1 \left(\frac{x}{2}\right)\right)$

Distribute the $\frac{1}{8}$ multiplier

$\frac{5 \cdot \ln | x - 2 |}{8} + \frac{5 \ln | x - 2 |}{16} - \frac{3 {\tan}^{-} 1 \left(\frac{x}{2}\right)}{8} + c$

Simplify:
$\frac{\left(5 - \ln | x - 2 |\right) - \left(3 \cdot {\tan}^{-} 1 \left(\frac{x}{2}\right)\right)}{8} + \frac{5 \ln | {x}^{2} + 4 |}{16} + c$

You can simplify further if you like, however this is an acceptable answer.