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

Mar 30, 2017

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

#### Explanation:

We want to find $A$, $B$, and $C$ such that $\frac{2 x + 1}{\left(x - 2\right) \left({x}^{2} + 4\right)} = \frac{A}{x - 2} + \frac{B x + C}{{x}^{2} + 4}$.

We multiply both sides by $\left(x - 2\right) \left({x}^{2} + 4\right)$ to get $2 x + 1 = A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 2\right)$. Since we want to find $A$, $B$, and $C$ such that all $x$ satisfies the above equation, we can arbitrarily substitute values that simplify the problem, such as $x = 2$.

Then, the equation would become $5 = 8 A$. $A$ then is $\frac{5}{8}$.

We can also substitute $x = 0$: $1 = 4 A - 2 C$. However, since $A = \frac{5}{8}$, $1 = \frac{5}{2} - 2 C$. Solving for $C$ gives $\frac{3}{4}$.

Knowing $A$ and $C$, we can easily solve for $B$ in this equation $2 x + 1 = \frac{5}{8} \left({x}^{2} + 4\right) + \left(B x + \frac{3}{4}\right) \left(x - 2\right)$ by setting $x = 1$ to get $B = - \frac{5}{8}$.

After substituting the values in and simplifying, we have successfully decomposed the fraction to $\frac{5}{8 \left(x - 2\right)} + \frac{- 5 x + 6}{8 \left({x}^{2} + 4\right)}$.

We can integrate the separate fractions one by one:

$\setminus \int \setminus \frac{5}{8 \left(x - 2\right)} \setminus \mathrm{dx}$
$= \frac{5}{8} \setminus \int \setminus \frac{1}{u} \setminus \mathrm{du}$ (substituting $u = x - 2$ and $\mathrm{du} = \mathrm{dx}$)
$= \frac{5}{8} \ln | u | + C$
$= \frac{5}{8} \ln | x - 2 | + C$ (substituting $u = x - 2$ back)

$\setminus \int \setminus \frac{- 5 x + 6}{8 \left({x}^{2} + 4\right)} \setminus \mathrm{dx}$
$= - \frac{5}{8} \setminus \int \setminus \frac{x - \frac{6}{5}}{{x}^{2} + 4} \setminus \mathrm{dx}$
$= - \frac{5}{8} \left(\setminus \int \setminus \frac{x}{{x}^{2} + 4} \setminus \mathrm{dx} - \setminus \int \setminus \frac{\frac{6}{5}}{{x}^{2} + 4} \setminus \mathrm{dx}\right)$
$= - \frac{5}{8} \left(\setminus \int \setminus \frac{x}{2 x u} \setminus \mathrm{du} - \setminus \int \setminus \frac{\frac{6}{5}}{{x}^{2} + 4} \setminus \mathrm{dx}\right)$ (substituting $u = {x}^{2} + 4$ and $\mathrm{du} = 2 x \setminus \mathrm{dx}$)
$= - \frac{5}{8} \left(\frac{1}{2} \ln | u | + C - \setminus \int \setminus \frac{\frac{6}{5}}{{x}^{2} + 4} \setminus \mathrm{dx}\right)$
$= - \frac{5}{8} \left(\frac{1}{2} \ln | {x}^{2} + 4 | + C - \setminus \int \setminus \frac{\frac{6}{5}}{{x}^{2} + 4} \setminus \mathrm{dx}\right)$ (substituting $u = {x}^{2} + 4$ back)
$= - \frac{5}{8} \left(\frac{1}{2} \ln | {x}^{2} + 4 | + C - \frac{6}{5} \setminus \int \setminus \frac{1}{{x}^{2} + 4} \setminus \mathrm{dx}\right)$
$= - \frac{5}{8} \left(\frac{1}{2} \ln | {x}^{2} + 4 | + C - \frac{6}{5} \setminus \int \setminus \frac{1}{4 \left({\tan}^{2} \left(\setminus \theta\right) + 1\right)} \setminus 2 {\sec}^{2} \left(\setminus \theta\right) \setminus d \setminus \theta\right)$ (substituting $\setminus 2 \tan \left(\setminus \theta\right) = x$ and $\mathrm{dx} = 2 {\sec}^{2} \left(\setminus \theta\right) \setminus d \setminus \theta$)
$= - \frac{5}{8} \left(\frac{1}{2} \ln | {x}^{2} + 4 | + C - \frac{6}{5} \setminus \int \setminus \frac{1}{4 {\sec}^{2} \left(\setminus \theta\right)} \setminus 2 {\sec}^{2} \left(\setminus \theta\right) \setminus d \setminus \theta\right)$
$= - \frac{5}{8} \left(\frac{1}{2} \ln | {x}^{2} + 4 | + C - \frac{3}{5} \setminus \theta\right)$
$= - \frac{5}{8} \left(\frac{1}{2} \ln | {x}^{2} + 4 | + C - \frac{3}{5} {\tan}^{-} 1 \left(\frac{x}{2}\right)\right)$ (since $\frac{x}{2} = \tan \left(\setminus \theta\right)$)
$= \frac{3}{8} {\tan}^{-} 1 \left(\frac{x}{2}\right) - \frac{5}{16} \ln | {x}^{2} + 4 | + C$

Finally, we add the two integrals to get the final answer: $\frac{5}{8} \ln | x - 2 | + \frac{3}{8} {\tan}^{-} 1 \left(\frac{x}{2}\right) - \frac{5}{16} \ln | {x}^{2} + 4 | + C$